4 Volume integrals involving vectors

Integrating a scalar function of a vector over a volume is essentially the same procedure as in HELM booklet  27.3. In 3D cartesian coordinates the volume element $dV$ is $dxdydz$ . The scalar function may be the divergence of a vector function.

Example 28

Integrate $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ over the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ where $\underset{̲}{F}$ is the vector function $x^{2}y\underset{̲}{i}+\left(x-z\right)\underset{̲}{j}+2x{z}^2\underset{̲}{k}$ .

Solution

$\underset{̲}{\nabla }\cdot \underset{̲}{F}=\frac{\partial }{\partial x}\left(x^{2}y\right)+\frac{\partial }{\partial y}\left(x-z\right)+\frac{\partial }{\partial z}\left(2x{z}^2\right)=2xy+4xz$

The integral is

Task!

Using spherical polar coordinates and the vector field $\underset{̲}{F}=r^2\underset{̲}{\stackrel{̂}{r}}+r^{2}sin\theta \underset{̲}{\stackrel{̂}{\mathrm{\&varphi;}}}$ , evaluate the integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$ over the sphere given by $r\le a$ .

Answer

Exercises

1. Evaluate $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$ when $\underset{̲}{F}$ is the vector field $yz\underset{̲}{i}+xy\underset{̲}{j}$ and $V$ is the unit cube $0\le x\le 1$ , $0\le y\le 1$ .
2. For the vector field $\underset{̲}{F}=\left(x^{2}y+sinz\right)\underset{̲}{i}+\left(x{y}^2+e^z\right)\underset{̲}{j}+\left(z^2+x^y\right)\underset{̲}{k}$ , find the integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$ where $V$ is the volume inside the tetrahedron bounded by $x=0$ , $y=0$ , $z=0$ and $x+y+z=1$ .

Answer

Less commonly, integrating a vector function over a volume integral is similar, but care should be taken with the various components. It may help to think in terms of a separate volume integral for each component. The vector function may be of the form $\underset{̲}{\nabla }f$ or $\underset{̲}{\nabla }\times \underset{̲}{F}$ .

Example 29

Integrate the function $\underset{̲}{F}=x^2\underset{̲}{i}+2\underset{̲}{j}$ over the prism given by $0\le x\le 1$ , $0\le y\le 2$ , $0\le z\le \left(1-x\right)$ . (See Figure 15.)

Figure 15:

Solution

The integral is

Example 30

For $\underset{̲}{F}=x^{2}y\underset{̲}{i}+y^2\underset{̲}{j}$ evaluate $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)dV$ where $V$ is the volume under the plane $z=x+y+2$ (and above $z=0$ ) for $-1\le x\le 1$ , $-1\le y\le 1$ .

Solution

$\underset{̲}{\nabla }\times \underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill x^{2}y\hfill & \hfill y^2\hfill & \hfill 0\hfill \end{array}\right|=-x^2\underset{̲}{k}$

so

Figure 16:

Task!

Evaluate the integral ${\int \; <!--nolimits\; -->}_V\underset{̲}{F}dV$ for the case where $\underset{̲}{F}=x\underset{̲}{i}+y^2\underset{̲}{j}+z\underset{̲}{k}$ and $V$ is the cube $-1\le x\le 1$ , $-1\le y\le 1$ , $-1\le z\le 1$ .

Answer

Exercises

1. For $f=x^2+yz$ , and $V$ the volume bounded by $y=0$ , $x+y=1$ and $-x+y=1$ for $-1\le z\le 1$ , find the integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left(\underset{̲}{\nabla }f\right)dV$ .
2. Evaluate the integral ${\int \; <!--nolimits\; -->}_V\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)dV$ for the case where $\underset{̲}{F}=xz\underset{̲}{i}+\left(x^3+y^3\right)\underset{̲}{j}-4y\underset{̲}{k}$ and $V$ is the cube $-1\le x\le 1$ , $-1\le y\le 1$ , $-1\le z\le 1$ .

Answer