Saydanzad Group Mathematical Methods in the Physical Sciences

Chapter 5

Multiple Integrals; Applications of Integration

5.1Single integral

Illustration of the definite integral as the signed area under a curve.
Figure 5.1. Illustration of the definite integral as the signed area under a curve.
Definition - Definite Integral as Signed Area

If \(f(x)\) is continuous on the closed interval \([a,b]\), the definite integral

\begin{equation} \int_a^b f(x)\,dx \end{equation}

represents the accumulated signed area between the curve \(y=f(x)\) and the \(x\)-axis, from \(x=a\) to \(x=b\).

  • Area above the \(x\)-axis contributes positive value.

  • Area below the \(x\)-axis contributes negative value.

5.2Double and Triple Integrals

In many physical problems, quantities depend on two or three spatial variables, e.g. temperature \(T(x,y,z)\), mass density \(\rho(x,y,z)\), or electric potential \(\Phi(x,y,z)\). To accumulate such quantities over a region in the plane or in space, we use multiple integrals.

Definition - Double Integral

Let \(f(x,y)\) be a continuous function on a region \(A \subset \mathbb{R}^2\). The double integral

\begin{equation} \iint_A f(x,y)\,dA \end{equation}

represents the accumulated value of \(f\) over the region \(A\).

  • If \(f(x,y) \ge 0\), it represents the volume under the surface \(z=f(x,y)\) above the region \(A\).

  • If \(f(x,y)\) takes negative values, the integral subtracts the area under the graph where \(f(x,y)<0\). This keeps the sign information of the function.

In particular, if \(f(x,y) \equiv 1\),

\begin{equation} \iint_A 1\,dA = \text{Area}(A). \end{equation}
Example

Find the volume of the solid below the plane \(z = 1 + y\), bounded by the coordinate planes and the vertical plane \(2x + y = 2\).

Geometry of the solid bounded by the coordinate planes, the vertical plane 2x + y = 2, and the plane z = 1 + y.
Figure 5.2. Geometry of the solid bounded by the coordinate planes, the vertical plane \(2x + y = 2\), and the plane \(z = 1 + y\).
Solution

Setup. The projection \(A\) of the solid onto the \(xy\)-plane is the triangle

\begin{equation} A=\{(x,y): x\ge0,\ y\ge0,\ 2x+y\le 2\}. \end{equation}

Since \(z=1+y\), the volume is

\begin{equation} V=\iint_A z\,dA=\iint_A (1+y)\,dx\,dy. \end{equation}

Method 1: integrate w.r.t. \(y\) first. For a fixed \(x\), \(y\) runs from \(0\) to \(2-2x\) and \(0\le x\le 1\). Thus

\begin{equation} \begin{aligned} V &=\int_{x=0}^{1}\!\left(\int_{y=0}^{2-2x} (1+y)\,dy\right)dx =\int_{0}^{1}\!\left[\,y+\frac{y^{2}}{2}\,\right]_{0}^{\,2-2x} dx \\ &=\int_{0}^{1}\!\Big((2-2x)+\tfrac12(2-2x)^2\Big)\,dx =\int_{0}^{1}\!\big(4-6x+2x^2\big)\,dx =\frac{5}{3}. \end{aligned} \end{equation}
Integrating y first. For a fixed x the strip runs from y=0 up to the line, and the element of area is A= y\, x.
Figure 5.3. Integrating \(y\) first. For a fixed \(x\) the strip runs from \(y=0\) up to the line, and the element of area is \(\Delta A=\Delta y\,\Delta x\).

Method 2: integrate w.r.t. \(x\) first. For a fixed \(y\), \(x\) runs from \(0\) to \(1-\tfrac{y}{2}\) and \(0\le y\le 2\). Hence

\begin{equation} \begin{aligned} V &=\int_{y=0}^{2}\!\left(\int_{x=0}^{\,1-y/2} (1+y)\,dx\right)dy =\int_{0}^{2} (1+y)\,\Big[x\Big]_{0}^{\,1-y/2}\,dy \\ &=\int_{0}^{2}\!(1+y)\!\left(1-\frac{y}{2}\right)dy =\int_{0}^{2}\!\left(1+\frac{y}{2}-\frac{y^{2}}{2}\right)dy =\frac{5}{3}. \end{aligned} \end{equation}
Integrating x first. For a fixed y the strip runs from x=0 to x=1-y/2, and the element of area is A= x\, y.
Figure 5.4. Integrating \(x\) first. For a fixed \(y\) the strip runs from \(x=0\) to \(x=1-y/2\), and the element of area is \(\Delta A=\Delta x\,\Delta y\).

5.3When to Integrate \(y\)-First vs. \(x\)-First

In evaluating a double integral, the choice of integration order depends on which limits are simpler or which inner integral is easier to evaluate. We typically choose the order that makes the region description and computation more convenient.

Type I Regions (Vertical Strips)

When the region \(A\) is bounded above and below by curves

\begin{equation} y = y_2(x), \qquad y = y_1(x), \end{equation}

and \(x\) varies between fixed limits \(a\) and \(b\), we integrate with respect to \(y\) first:

\begin{equation} \iint_A f(x,y)\,dx\,dy = \int_{x=a}^{b} \!\! \left( \int_{y=y_1(x)}^{y_2(x)} f(x,y)\,dy \right) dx. \label{eq:typeI} \end{equation}
Type I region A: vertical slices, integrating y first.
Figure 5.5. Type I region \(A\): vertical slices, integrating \(y\) first.

Type II Regions (Horizontal Strips)

When the region \(A\) is bounded on the left and right by curves

\begin{equation} x = x_1(y), \qquad x = x_2(y), \end{equation}

and \(y\) varies between fixed limits \(c\) and \(d\), we integrate with respect to \(x\) first:

\begin{equation} \iint_A f(x,y)\,dx\,dy = \int_{y=c}^{d} \!\! \left( \int_{x=x_1(y)}^{x_2(y)} f(x,y)\,dx \right) dy. \label{eq:typeII} \end{equation}
Type II region A: horizontal slices, integrating x first.
Figure 5.6. Type II region \(A\): horizontal slices, integrating \(x\) first.

Either Description Works

If the region \(A\) satisfies both descriptions, either order of integration yields the same result:

\begin{equation} \iint_A f(x,y)\,dx\,dy = \int_{x=a}^{b} \!\! \left( \int_{y=y_1(x)}^{y_2(x)} f(x,y)\,dy \right) dx = \int_{y=c}^{d} \!\! \left( \int_{x=x_1(y)}^{x_2(y)} f(x,y)\,dx \right) dy. \label{eq:eitherorder} \end{equation}
Regions where either order is valid.
Figure 5.7. Regions where either order is valid.
Remark - Product Integrands over Rectangular Regions

If the region \(A = [a,b] \times [c,d]\) is a rectangle and the integrand separates as \(f(x,y) = g(x)\,h(y)\), then

\begin{equation} \iint_A f(x,y)\,dx\,dy = \left( \int_a^b g(x)\,dx \right) \left( \int_c^d h(y)\,dy \right). \end{equation}

This property can greatly simplify double-integral evaluations.

Definition - Triple integral

For a continuous \(f\) on a solid \(R\subset\mathbb{R}^3\),

\begin{equation} \iiint_R f(x,y,z)\,dV \end{equation}

is the accumulated signed total of \(f\) throughout the solid \(R\). (If \(f\equiv 1\), it equals the volume of \(R\); if \(f=\rho\) is mass density, it equals the mass.)

Example

Find the volume of the solid bounded by the coordinate planes, the vertical plane \(2x+y=2\), and the plane \(z=1+y\) via a triple integral

Solution.

Region and bounds. Projecting onto the \(xy\)-plane gives the triangle

\begin{equation} A=\{(x,y):\ 0\le x\le 1,\ \ 0\le y\le 2-2x\}. \end{equation}

For each \((x,y)\in A\), \(z\) ranges from the bottom \(z=0\) to the top \(z=1+y\).

Triple–integral setup.

\begin{equation} dV = dx\,dy\,dz\end{equation}

as a result:

\begin{equation} V=\iiint_{V} \,dV =\int_{x=0}^{1}\int_{y=0}^{\,2-2x}\int_{z=0}^{\,1+y} \,dz\,dy\,dx. \end{equation}

Evaluate (inner integral first).

\begin{equation} \int_{z=0}^{\,1+y} 1\,dz = 1+y, \qquad\Rightarrow\qquad V=\int_{0}^{1}\int_{0}^{\,2-2x} (1+y)\,dy\,dx. \end{equation}

The remaining double integral is elementary:

\begin{equation} \int_{0}^{1}\!\left[\,y+\frac{y^{2}}{2}\,\right]_{0}^{\,2-2x}\!dx =\int_{0}^{1} \big(4-6x+2x^{2}\big)\,dx =\boxed{\ \dfrac{5}{3}\ }. \end{equation}
Example

On the same solid \(V\) as above, suppose the density (mass per unit volume) is \(\rho(x,y,z)=x+z\). Find the total mass \(M\).

Solution.

Triple–integral setup.

\begin{equation} dM = \rho(x,y,z)\,dx\,dy\,dz\end{equation}

as a result:

\begin{equation} M=\iiint_{V} \rho(x,y,z)\,dV =\int_{x=0}^{1}\int_{y=0}^{\,2-2x}\int_{z=0}^{\,1+y} (x+z)\,dz\,dy\,dx. \end{equation}

Step 1. Integrate in \(z\).

\begin{equation} \int_{0}^{\,1+y} (x+z)\,dz =\Big[xz+\tfrac{z^{2}}{2}\Big]_{0}^{\,1+y} =x(1+y)+\frac{(1+y)^{2}}{2}. \end{equation}

Step 2. Integrate in \(y\), then \(x\).

\begin{equation} M=\int_{0}^{1}\int_{0}^{\,2-2x}\!\!\left(x(1+y)+\frac{(1+y)^{2}}{2}\right)dy\,dx =\int_{0}^{1}\!\left(\frac{2x^{3}}{3}-5x+\frac{13}{3}\right)\!dx =\boxed{\ 2\ }. \end{equation}

5.4Applications of Integration

Center of mass

Definition - Center of mass

For any mass distribution with mass element \(dm\) and position vector \(\vec r = x\,\hat{\mathbf i}+y\,\hat{\mathbf j}+z\,\hat{\mathbf k}\), the center of mass is

\begin{equation} \boxed{\quad \vec r_{\mathrm{cm}}=\frac{1}{M}\int \vec r\,dm, \qquad M=\int dm \quad }. \end{equation}

Specializations (same formula, different \(dm\)):

\begin{equation} \begin{aligned} &\text{Surface (lamina):}\quad dm=\sigma(x,y)\,dA \ \Rightarrow\ M=\iint \sigma\,dA,\quad \vec r_{\mathrm{cm}}=\frac{1}{M}\iint \vec r\,\sigma\,dA.\\[4pt] &\text{Volume (solid):}\quad dm=\rho(x,y,z)\,dV \ \Rightarrow\ M=\iiint \rho\,dV,\quad \vec r_{\mathrm{cm}}=\frac{1}{M}\iiint \vec r\,\rho\,dV. \end{aligned} \end{equation}

Components: \(x_{\mathrm{cm}}=\dfrac{1}{M}\int x\,dm,\ y_{\mathrm{cm}}=\dfrac{1}{M}\int y\,dm,\ z_{\mathrm{cm}}=\dfrac{1}{M}\int z\,dm.\)

Right triangular lamina with vertices (0,0), (L,0), (0,H).
Figure 5.8. Right triangular lamina with vertices \((0,0)\), \((L,0)\), \((0,H)\).
Example

Centroid of a Right Triangle (Uniform Density \(\sigma_0\))

Solution.

Step 1. Describe the region. The hypotenuse runs from \((L,0)\) to \((0,H)\), so its equation is \(\dfrac{x}{L}+\dfrac{y}{H}=1\), that is \(y = H\!\left(1-\dfrac{x}{L}\right)\). A convenient Type I description is therefore

\begin{equation} R=\Big\{(x,y):\ 0\le x\le L,\quad 0\le y\le H\!\left(1-\frac{x}{L}\right)\Big\}. \end{equation}

Step 2. Mass.

\begin{equation} \begin{aligned} M &=\iint_R \sigma_0\,dA =\sigma_0\int_{0}^{L}\!\!\int_{0}^{\,H(1-x/L)} dy\,dx =\sigma_0\int_{0}^{L}\!\left(H-\frac{H}{L}x\right)dx =\sigma_0\,\frac{HL}{2}. \end{aligned} \end{equation}

Step 3. First moments.

\begin{equation} \begin{aligned} \iint_R x\,\sigma_0\,dA &=\sigma_0\int_{0}^{L}\!\!\int_{0}^{\,H(1-x/L)} x\,dy\,dx =\sigma_0\int_{0}^{L} x\!\left(H-\frac{H}{L}x\right)dx =\sigma_0\,\frac{HL^2}{6},\\[4pt] \iint_R y\,\sigma_0\,dA &=\sigma_0\int_{0}^{L}\!\!\int_{0}^{\,H(1-x/L)} y\,dy\,dx =\sigma_0\int_{0}^{L}\frac{1}{2}\!\left(H-\frac{H}{L}x\right)^{\!2}\,dx =\sigma_0\,\frac{H^2L}{6}. \end{aligned} \end{equation}

Step 4. Coordinates of the center of mass.

\begin{equation} x_{cm}=\frac{1}{M}\iint_R x\,\sigma_0\,dA =\frac{\sigma_0(HL^2/6)}{\sigma_0(HL/2)}=\frac{L}{3},\qquad y_{cm}=\frac{1}{M}\iint_R y\,\sigma_0\,dA =\frac{\sigma_0(H^2L/6)}{\sigma_0(HL/2)}=\frac{H}{3}. \end{equation}

Result. The centroid is

\begin{equation} \boxed{(x_{cm},y_{cm})=\left(\dfrac{L}{3},\,\dfrac{H}{3}\right)}. \end{equation}

5.5Change of Variables in Integrals

Cartesian coordinates are rarely the natural ones for a physical region. A cylinder, a sphere or a cone has boundaries that are awkward to describe with \(x\), \(y\) and \(z\) but trivial in the right coordinates: a sphere of radius \(a\) is simply \(r = a\). The price is that the volume element is no longer \(dx\,dy\,dz\), and the next two subsections derive what it becomes.

Change of Variables: Cartesian \(\rightarrow\) Cylindrical Coordinates

1. Definition of Cylindrical Coordinates

A point \(P(x, y, z)\) in space can be described using cylindrical coordinates \((\rho, \phi, z)\) defined as:

\begin{equation} \begin{cases} x = \rho \cos\phi, \\[4pt] y = \rho \sin\phi, \\[4pt] z = z. \end{cases} \end{equation}

Here:

The position vector of the point is

\begin{equation} \vec{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}. \end{equation}

Substituting the cylindrical coordinate relations:

\begin{equation} \vec{r} = \rho(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}) + z\,\hat{k}. \end{equation}

Thus, we can write

\begin{equation} \boxed{\vec{r} = \rho\,\hat{\rho} + z\,\hat{k}}, \end{equation}

where \(\hat{\rho}\) is the radial unit vector defined by

\begin{equation} \boxed{ \hat{\rho} = \cos\phi\,\hat{i} + \sin\phi\,\hat{j}. } \end{equation}

2. The Local Unit Vectors

At each point, three mutually perpendicular unit vectors are defined:

\begin{equation} \begin{aligned} \hat{\rho} &= \cos\phi\,\hat{i} + \sin\phi\,\hat{j}, \\[4pt] \hat{\phi} &= -\sin\phi\,\hat{i} + \cos\phi\,\hat{j}, \\[4pt] \hat{k} &= \hat{k}. \end{aligned} \end{equation}

These three vectors form a right-handed orthonormal basis:

\begin{equation} \hat{\rho}\times\hat{\phi} = \hat{k}, \qquad \hat{\phi}\times\hat{k} = \hat{\rho}, \qquad \hat{k}\times\hat{\rho} = \hat{\phi}. \end{equation}

3. Differential Displacement \(d\vec{r}\)

Starting from

\begin{equation} \vec{r} = \rho\,\hat{\rho} + z\,\hat{k}, \end{equation}

we differentiate:

\begin{equation} d\vec{r} = d\rho\,\hat{\rho} + \rho\,d\hat{\rho} + dz\,\hat{k}. \end{equation}

Since \(\hat{\rho}\) depends only on \(\phi\),

\begin{equation} d\hat{\rho} = \frac{\partial \hat{\rho}}{\partial \phi}\,d\phi. \end{equation}

Compute this derivative:

\begin{equation} \frac{\partial \hat{\rho}}{\partial \phi} = -\sin\phi\,\hat{i} + \cos\phi\,\hat{j} = \hat{\phi}. \end{equation}

Hence,

\begin{equation} d\hat{\rho} = \hat{\phi}\,d\phi. \end{equation}

Substitute into \(d\vec{r}\):

\begin{equation} \boxed{ d\vec{r} = \hat{\rho}\,d\rho + \rho\,\hat{\phi}\,d\phi + \hat{k}\,dz. } \end{equation}
Differential volume element in cylindrical coordinates.
Figure 5.9. Differential volume element in cylindrical coordinates.

4. Small Surface Element \(d\vec{s}\)

Each coordinate surface has a small area element with magnitude and direction as follows:

\begin{equation} \begin{aligned} d\vec{s}_\rho &= (\rho\, d\phi\, \hat{\phi}) \times (dz\, \hat{k}) = (\rho\, d\phi\, dz)\, \hat{\rho}, && \text{(surface of a cylinder, constant $\rho$)}, \\[6pt] d\vec{s}_\phi &= (dz\, \hat{k}) \times (d\rho\, \hat{\rho}) = (d\rho\, dz)\, \hat{\phi}, && \text{(surface of a plane at constant $\phi$)}, \\[6pt] d\vec{s}_z &= (d\rho\, \hat{\rho}) \times (\rho\, d\phi\, \hat{\phi}) = (\rho\, d\rho\, d\phi)\, \hat{k}, && \text{(surface of a disk, constant $z$)}. \end{aligned} \end{equation}
\begin{equation} \boxed{ d\vec{s}_\rho = \rho\, d\phi\, dz\, \hat{\rho}, \qquad d\vec{s}_\phi = d\rho\, dz\, \hat{\phi}, \qquad d\vec{s}_z = \rho\, d\rho\, d\phi\, \hat{k}. } \end{equation}

5. Small Volume Element \(dV\)

The infinitesimal volume element is formed by the product of the three small lengths:

\begin{equation} d\rho, \quad \rho\,d\phi, \quad dz. \end{equation}

Thus,

\begin{equation} \boxed{ dV = \rho\, d\rho\, d\phi\, dz. } \end{equation}
QuantityExpressionDescription
\(\vec{r}\)\(\rho\,\hat{\rho} + z\,\hat{k}\)Position vector
\(d\vec{r}\)\(\hat{\rho}\,d\rho + \rho\,\hat{\phi}\,d\phi + \hat{k}\,dz\)Differential displacement
\(d\vec{s}_\rho\)\(\rho\,d\phi\,dz\,\hat{\rho}\)Surface element on cylindrical wall
\(d\vec{s}_\phi\)\(d\rho\,dz\,\hat{\phi}\)Surface element on plane at constant \(\phi\)
\(d\vec{s}_z\)\(\rho\,d\rho\,d\phi\,\hat{k}\)Surface element on disk (constant \(z\))
\(dV\)\(\rho\,d\rho\,d\phi\,dz\)Volume element
Table 5.1. Summary of differential elements in cylindrical coordinates

Change of Variables: Cartesian \(\rightarrow\) Spherical Coordinates

1. Definition of Spherical Coordinates

A point \(P(x,y,z)\) in space can be described using spherical coordinates \((r, \theta, \phi)\) defined as:

\begin{equation} \begin{cases} x = r \sin\theta \cos\phi, \\[4pt] y = r \sin\theta \sin\phi, \\[4pt] z = r \cos\theta. \end{cases} \end{equation}

Here:

The position vector of the point is

\begin{equation} \vec{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}. \end{equation}

Substituting the spherical coordinate relations:

\begin{equation} \vec{r} = r(\sin\theta\cos\phi\,\hat{i} + \sin\theta\sin\phi\,\hat{j} + \cos\theta\,\hat{k}). \end{equation}

Thus, we can write

\begin{equation} \boxed{\vec{r} = r\,\hat{r}}, \end{equation}

where \(\hat{r}\) is the radial unit vector defined by

\begin{equation} \boxed{ \hat{r} = \sin\theta\cos\phi\,\hat{i} + \sin\theta\sin\phi\,\hat{j} + \cos\theta\,\hat{k}. } \end{equation}

2. The Local Unit Vectors

At each point, three mutually perpendicular unit vectors are defined:

\begin{equation} \begin{aligned} \hat{r} &= \sin\theta\cos\phi\,\hat{i} + \sin\theta\sin\phi\,\hat{j} + \cos\theta\,\hat{k}, \\[4pt] \hat{\theta} &= \cos\theta\cos\phi\,\hat{i} + \cos\theta\sin\phi\,\hat{j} - \sin\theta\,\hat{k}, \\[4pt] \hat{\phi} &= -\sin\phi\,\hat{i} + \cos\phi\,\hat{j}. \end{aligned} \end{equation}

These three vectors form a right-handed orthonormal basis:

\begin{equation} \hat{r}\times\hat{\theta} = \hat{\phi}, \qquad \hat{\theta}\times\hat{\phi} = \hat{r}, \qquad \hat{\phi}\times\hat{r} = \hat{\theta}. \end{equation}

3. Differential Displacement \(d\vec{r}\)

Differentiating the position vector

\begin{equation} \vec{r} = r\,\hat{r}, \end{equation}

we obtain

\begin{equation} d\vec{r} = dr\,\hat{r} + r\,d\hat{r}. \end{equation}

To find \(d\hat{r}\), note that \(\hat{r}\) depends on both \(\theta\) and \(\phi\):

\begin{equation} d\hat{r} = \frac{\partial \hat{r}}{\partial \theta} d\theta + \frac{\partial \hat{r}}{\partial \phi} d\phi. \end{equation}

Compute these derivatives:

\begin{equation} \frac{\partial \hat{r}}{\partial \theta} = \hat{\theta}, \qquad \frac{\partial \hat{r}}{\partial \phi} = \sin\theta\,\hat{\phi}. \end{equation}

Hence,

\begin{equation} d\hat{r} = \hat{\theta}\,d\theta + \sin\theta\,\hat{\phi}\,d\phi. \end{equation}

Substitute into \(d\vec{r} = dr\,\hat{r} + r\,d\hat{r}\):

\begin{equation} \boxed{ d\vec{r} = \hat{r}\,dr + r\,\hat{\theta}\,d\theta + r\sin\theta\,\hat{\phi}\,d\phi. } \end{equation}
Differential displacement in spherical coordinates
Figure 5.10. Differential displacement in spherical coordinates

4. Small Surface Element \(d\vec{s}\)

Each coordinate surface has a small area element with magnitude and direction as follows:

\begin{equation} \begin{aligned} d\vec{s}_r &= (r\, d\theta\, \hat{\theta}) \times (r\sin\theta\, d\phi\, \hat{\phi}) = (r^2 \sin\theta\, d\theta\, d\phi)\, \hat{r}, && \text{(surface of a sphere, constant $r$)}, \\[6pt] d\vec{s}_\theta &= (r\sin\theta\, d\phi\, \hat{\phi}) \times (dr\, \hat{r}) = (r\sin\theta\, dr\, d\phi)\, \hat{\theta}, && \text{(surface of a cone, constant $\theta$)}, \\[6pt] d\vec{s}_\phi &= (dr\, \hat{r}) \times (r\, d\theta\, \hat{\theta}) = (r\, dr\, d\theta)\, \hat{\phi}, && \text{(surface of a half-plane, constant $\phi$)}. \end{aligned} \end{equation}
\begin{equation} \boxed{ d\vec{s}_r = r^2 \sin\theta\, d\theta\, d\phi\, \hat{r}, \qquad d\vec{s}_\theta = r \sin\theta\, dr\, d\phi\, \hat{\theta}, \qquad d\vec{s}_\phi = r\, dr\, d\theta\, \hat{\phi}. } \end{equation}

5. Small Volume Element \(dV\)

The infinitesimal volume is formed by the product of the three differential lengths:

\begin{equation} dr, \quad r\,d\theta, \quad r\sin\theta\,d\phi. \end{equation}

Thus,

\begin{equation} \boxed{ dV = r^2 \sin\theta\, dr\, d\theta\, d\phi. } \end{equation}
QuantityExpressionDescription
\(\vec{r}\)\(r\,\hat{r}\)Position vector
\(d\vec{r}\)\(\hat{r}\,dr + r\,\hat{\theta}\,d\theta + r\sin\theta\,\hat{\phi}\,d\phi\)Differential displacement
\(d\vec{s}_r\)\(r^2\sin\theta\, d\theta\, d\phi\,\hat{r}\)Surface element on sphere (\(r=\text{const}\))
\(d\vec{s}_\theta\)\(r\sin\theta\, dr\, d\phi\,\hat{\theta}\)Surface element on cone (\(\theta=\text{const}\))
\(d\vec{s}_\phi\)\(r\,dr\, d\theta\,\hat{\phi}\)Surface element on half-plane (\(\phi=\text{const}\))
\(dV\)\(r^2 \sin\theta\, dr\, d\theta\, d\phi\)Volume element
Table 5.2. Summary of differential elements in spherical coordinates

5.6Connection with the Jacobian Matrix

When transforming a multiple integral from one set of variables to another, the Jacobian determinant accounts for the change in scale between coordinate systems.

Theorem - Change of Variables for Triple Integrals

Suppose we have a triple integral

\begin{equation} \iiint f(u,v,w)\,du\,dv\,dw \end{equation}

in some variables \(u, v, w\). Let \(r, s, t\) be another set of variables related by

\begin{equation} u = u(r,s,t), \qquad v = v(r,s,t), \qquad w = w(r,s,t). \end{equation}

as a result:

\begin{equation} \begin{aligned} du &= \frac{\partial u}{\partial r}\,dr + \frac{\partial u}{\partial s}\,ds + \frac{\partial u}{\partial t}\,dt,\\[4pt] dv &= \frac{\partial v}{\partial r}\,dr + \frac{\partial v}{\partial s}\,ds + \frac{\partial v}{\partial t}\,dt,\\[4pt] dw &= \frac{\partial w}{\partial r}\,dr + \frac{\partial w}{\partial s}\,ds + \frac{\partial w}{\partial t}\,dt. \end{aligned} \end{equation}

In matrix (differential) form,

\begin{equation} \begin{pmatrix} du\\ dv\\ dw \end{pmatrix} = \underbrace{ \begin{pmatrix} \frac{\partial u}{\partial r} & \frac{\partial u}{\partial s} & \frac{\partial u}{\partial t}\\[4pt] \frac{\partial v}{\partial r} & \frac{\partial v}{\partial s} & \frac{\partial v}{\partial t}\\[4pt] \frac{\partial w}{\partial r} & \frac{\partial w}{\partial s} & \frac{\partial w}{\partial t} \end{pmatrix} }_{J(\frac{u,v,w}{r,s,t})} \begin{pmatrix} dr\\ ds\\ dt \end{pmatrix}. \end{equation}

where \(J\) is the Jacobian of \(u,v,w\) with respect to \(r,s,t\), then the triple integral in the new variables becomes

\begin{equation} \iiint f(u,v,w)\,du\,dv\,dw = \iiint f\big(u(r,s,t),\,v(r,s,t),\,w(r,s,t)\big)\,\left|J\right|\,dr\,ds\,dt. \end{equation}

The integrand on the right is the same function \(f\), evaluated at the point the new coordinates describe; it is not a new function of \(r,s,t\). The factor \(\left|J\right|\) is what corrects for the fact that a box \(dr\,ds\,dt\) in the new coordinates does not have the same volume as a box \(du\,dv\,dw\) in the old ones.

Remark

The function \(f\) and the determinant \(J\) must both be expressed in the new variables, and the limits of integration must be adjusted to correspond to the transformed region.

Verifying the Jacobian for Cylindrical Coordinates

For the transformation \((\rho, \phi, z) \to (x, y, z)\),

\begin{equation} \begin{cases} x = \rho\cos\phi, \\[4pt] y = \rho\sin\phi, \\[4pt] z = z. \end{cases} \end{equation}

The Jacobian matrix is

\begin{equation} J\!\left(\frac{x,y,z}{\rho,\phi,z}\right) = \begin{pmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial z} \\[4pt] \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial z} \\[4pt] \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial z} \end{pmatrix} = \begin{pmatrix} \cos\phi & -\rho\sin\phi & 0 \\[4pt] \sin\phi & \ \rho\cos\phi & 0 \\[4pt] 0 & 0 & 1 \end{pmatrix}. \end{equation}

Compute its determinant:

\begin{equation} \det(J) = \begin{vmatrix} \cos\phi & -\rho\sin\phi & 0 \\[4pt] \sin\phi & \ \rho\cos\phi & 0 \\[4pt] 0 & 0 & 1 \end{vmatrix} = \rho. \end{equation}
Remark - Cylindrical Volume Element
\begin{equation} dV = \left|J\right|\,d\rho\,d\phi\,dz \;=\; \rho\,d\rho\,d\phi\,dz. \end{equation}

Verifying the Jacobian for Spherical Coordinates

For the transformation \((r, \theta, \phi) \to (x, y, z)\),

\begin{equation} \begin{cases} x = r\sin\theta\cos\phi, \\[4pt] y = r\sin\theta\sin\phi, \\[4pt] z = r\cos\theta. \end{cases} \end{equation}

The Jacobian matrix is

\begin{equation} J\!\left(\frac{x,y,z}{r,\theta,\phi}\right) = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\[4pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\[4pt] \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{pmatrix}. \end{equation}

Compute each partial derivative:

\begin{equation} \begin{aligned} \frac{\partial x}{\partial r} &= \sin\theta\cos\phi, & \frac{\partial x}{\partial \theta} &= r\cos\theta\cos\phi, & \frac{\partial x}{\partial \phi} &= -r\sin\theta\sin\phi, \\[4pt] \frac{\partial y}{\partial r} &= \sin\theta\sin\phi, & \frac{\partial y}{\partial \theta} &= r\cos\theta\sin\phi, & \frac{\partial y}{\partial \phi} &= r\sin\theta\cos\phi, \\[4pt] \frac{\partial z}{\partial r} &= \cos\theta, & \frac{\partial z}{\partial \theta} &= -r\sin\theta, & \frac{\partial z}{\partial \phi} &= 0. \end{aligned} \end{equation}

Therefore,

\begin{equation} \det(J) = \begin{vmatrix} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi \\[4pt] \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi \\[4pt] \cos\theta & -r\sin\theta & 0 \end{vmatrix}. \end{equation}

Expanding the determinant gives

\begin{equation} |J| = r^2\sin\theta. \end{equation}
Remark - Spherical Volume Element
\begin{equation} dV = \left|J\right|\,dr\,d\theta\,d\phi = r^2\sin\theta\,dr\,d\theta\,d\phi. \end{equation}
Example

Centroid and Moment of Inertia of a Uniform Solid Cone. Find the \(z\)-coordinate of the centroid \(z_{cm}\) of a uniform solid right circular cone (one nappe) whose height and base radius are both equal to \(h\), that is, one with semi-vertical angle \(45^\circ\). Also find the moment of inertia \(I_z\) about its symmetry axis.

The uniform solid cone of height h and base radius r=h, with its vertex at the origin and its symmetry axis along z.
Figure 5.11. The uniform solid cone of height \(h\) and base radius \(r=h\), with its vertex at the origin and its symmetry axis along \(z\).
Solution

Geometry and coordinates. Place the apex at the origin and the axis along \(+z\). The semi-vertical angle is \(45^\circ\), so the base radius equals the height: both are \(h\). That is why the radius of the cross-section at height \(z\) is itself \(z\), and the lateral surface is \(\rho=z\) (for \(0\le z\le h\)). Use cylindrical coordinates with \(dV=\rho\,d\rho\,d\phi\,dz\) and constant density \(\sigma\). Note that \(\sigma\) denotes the volume density here, written this way because \(\rho\) is already in use as the cylindrical radial coordinate.

\begin{equation} 0\le \phi\le 2\pi,\qquad 0\le \rho\le z,\qquad 0\le z\le h. \end{equation}

Step1: Mass.

\begin{equation} \begin{aligned} M&=\iiint \sigma\,dV = \sigma\int_{z=0}^{h}\int_{\rho=0}^{z}\int_{\phi=0}^{2\pi} \rho\,d\phi\,d\rho\,dz = \sigma\cdot 2\pi\int_{0}^{h}\!\!\left[\frac{\rho^{2}}{2}\right]_{0}^{z}dz \\ &= \sigma\pi\int_{0}^{h} z^{2}\,dz = \sigma\pi\left[\frac{z^{3}}{3}\right]_{0}^{h} = \boxed{\;\dfrac{\sigma\pi h^{3}}{3}\; }. \end{aligned} \end{equation}

Step2: Centroid \( z_{cm}\). Use \( z_{cm}=\dfrac{1}{M}\iiint z\,\sigma\,dV\).

\begin{equation} \begin{aligned} \iiint z\,\sigma\,dV &= \sigma\int_{0}^{h}\int_{0}^{z}\int_{0}^{2\pi} z\,\rho\,d\phi\,d\rho\,dz = \sigma\cdot 2\pi\int_{0}^{h} z\!\left[\frac{\rho^{2}}{2}\right]_{0}^{z}\!dz \\ &= \sigma\pi\int_{0}^{h} z^{3}\,dz = \sigma\pi\left[\frac{z^{4}}{4}\right]_{0}^{h} = \frac{\sigma\pi h^{4}}{4}. \end{aligned} \end{equation}

Hence

\begin{equation} z_{cm}=\frac{\sigma\pi h^{4}/4}{\sigma\pi h^{3}/3} =\boxed{\;\frac{3}{4}\,h\; }. \end{equation}

Step3: Moment of inertia about the axis. \(I_z=\iiint \rho^{2}\,dm=\sigma\iiint \rho^{2}\,dV\).

\begin{equation} \begin{aligned} I_z &= \sigma\int_{0}^{h}\int_{0}^{z}\int_{0}^{2\pi} \rho^{2}\,(\rho\,d\phi\,d\rho\,dz) = \sigma\cdot 2\pi\int_{0}^{h}\!\left[\frac{\rho^{4}}{4}\right]_{0}^{z}dz \\ &= \frac{\sigma\pi}{2}\int_{0}^{h} z^{4}\,dz = \frac{\sigma\pi}{2}\left[\frac{z^{5}}{5}\right]_{0}^{h} = \boxed{\;\frac{\sigma\pi h^{5}}{10}\; }. \end{aligned} \end{equation}

In terms of \(M\),

\begin{equation} I_z=\frac{\sigma\pi h^{5}}{10} =\frac{3}{10}\left(\frac{\sigma\pi h^{3}}{3}\right)h^{2} =\boxed{\;\frac{3}{10}\,M\,h^{2}\; }. \end{equation}
Example

Evaluate

\begin{equation} \int_{y=0}^{1}\!\int_{x=0}^{\,1-y} \exp\!\left(\frac{x-y}{x+y}\right)\,dx\,dy \end{equation}

by the substitution \( u=x-y,\quad v=x+y. \)

Solution.

Step 1. Map and Jacobian. The inverse relations are

\begin{equation} x=\frac{u+v}{2},\qquad y=\frac{v-u}{2}. \end{equation}

Hence

\begin{equation} J\!\left(\frac{x,y}{u,v}\right) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\[4pt] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac12 & \frac12\\[4pt] -\frac12 & \frac12 \end{vmatrix} =\frac12 \quad\Rightarrow\quad dx\,dy=\frac12\,du\,dv. \end{equation}

Step 2. Region in \((u,v)\). The original region is \( R=\{(x,y): x\ge 0,\ y\ge 0,\ x+y\le 1\}. \) On \(R\),

\begin{equation} v=x+y\in[0,1],\qquad u=x-y\in[-v,\,v] \end{equation}

(because along \(x=0\Rightarrow u=-v\) and along \(y=0\Rightarrow u=v\)).

Step 3. Transform the integrand and limits.

\begin{equation} \frac{x-y}{x+y}=\frac{u}{v},\qquad \iint_R e^{(x-y)/(x+y)}\,dx\,dy =\int_{v=0}^{1}\!\int_{u=-v}^{v} e^{u/v}\,\frac12\,du\,dv. \end{equation}

Step 4. Integrate. For fixed \(v>0\),

\begin{equation} \int_{-v}^{v} e^{u/v}\,du = v\,[e^{u/v}]_{-v}^{v} = v\,(e- e^{-1}) = 2v\,\sinh(1). \end{equation}

Therefore

\begin{equation} \int_{0}^{1}\!\int_{-v}^{v} \frac12\,e^{u/v}\,du\,dv =\frac12\int_{0}^{1} 2v\,\sinh(1)\,dv =\sinh(1)\!\int_{0}^{1} v\,dv =\frac12\,\sinh(1). \end{equation}
\begin{equation} \boxed{\displaystyle \int_{0}^{1}\!\int_{0}^{1-y} \exp\!\left(\frac{x-y}{x+y}\right)\,dx\,dy = \frac12\,\sinh(1)} \end{equation}

5.7Where Multiple Integrals Are Used

Every integral in this chapter is an accumulation: chop a region into pieces, evaluate a quantity on each piece, and add. What changes from one application to the next is only what sits inside the integral.

Remark

Two habits are worth taking from this chapter. First, draw the region and decide the order of integration before writing anything down; most of the difficulty in a multiple integral is in the limits, not the antiderivatives. Second, choose coordinates that match the symmetry of the region. A sphere described in Cartesian coordinates produces limits containing \(\sqrt{a^2-x^2-y^2}\); in spherical coordinates the same region is just \(r \le a\), and the price is the single factor \(r^2\sin\theta\).

Summary Table

ItemExpressionNotes
Double integral\(\iint_A f\,dA\)Volume under \(z=f(x,y)\); area if \(f\equiv1\).
Triple integral\(\iiint_R f\,dV\)Volume if \(f\equiv1\); mass if \(f=\rho\).
Type I region\(\int_a^b\!\int_{y_1(x)}^{y_2(x)} f\,dy\,dx\)Vertical strips; \(y\) first.
Type II region\(\int_c^d\!\int_{x_1(y)}^{x_2(y)} f\,dx\,dy\)Horizontal strips; \(x\) first.
Separable integrand\(\left(\int_a^b g\right)\left(\int_c^d h\right)\)Only on a rectangle, with \(f=g(x)h(y)\).
Centre of mass\(\vec r_{\mathrm{cm}} = \frac{1}{M}\int \vec r\,dm\)\(dm = \sigma\,dA\) or \(\rho\,dV\).
Cylindrical\(x=\rho\cos\phi\), \(y=\rho\sin\phi\), \(z=z\)\(dV = \rho\,d\rho\,d\phi\,dz\).
Spherical\(x=r\sin\theta\cos\phi\), , \(z=r\cos\theta\)\(dV = r^2\sin\theta\,dr\,d\theta\,d\phi\).
Jacobian\(J = \det\!\big(\frac{\partial (u,v,w)}{\partial (r,s,t)}\big)\)\(du\,dv\,dw = \left|J\right|\,dr\,ds\,dt\).
Cylindrical Jacobian\(\left|J\right| = \rho\)Recovers \(dV = \rho\,d\rho\,d\phi\,dz\).
Spherical Jacobian\(\left|J\right| = r^2\sin\theta\)Recovers \(dV = r^2\sin\theta\,dr\,d\theta\,d\phi\).

For the Interested Reader

The hardest part of a multiple integral is almost never the antiderivative — it is setting up the limits and choosing the coordinates. The sources below have far more practice at exactly that than we have room for. Everything listed is free.

Setting up double and triple integrals

Cylindrical, spherical and the Jacobian

Remark

We derived the volume elements twice on purpose: once geometrically, by multiplying the three small lengths along the coordinate directions, and once algebraically, as the Jacobian determinant. The geometric route shows why the factors \(\rho\) and \(r^2\sin\theta\) appear; the Jacobian route is the one that generalises to coordinates with no easy picture. It is worth being comfortable with both, because physics arguments usually quote the first and calculations usually use the second.