verifying the divergence theorem for the region in $R^3$ bounded by the surfaces $z=1-x^2$, $y=0$, $y=1$ and the x-y plane

Question

I am stuck on the following question. Vector calculus is not a forte of mine.

Let V be the region in $R^3$ bounded by the surfaces $z=1-x^2$, $y=0$, $y=1$ and the x-y plane. S is the closed surface of V with outward orientation from the solid, and let n deonte the unit normal vector in the direction of the orientation. Consider the vector field $$F(x,y,z) = (z^2-x)i-xyj+3zk$$ Verify the result of Gauss Divergence Theorem holds for this case. That is show that the surface integral and the triple integral in the theorem evaluate to the same number.

Thanks for your help :)

Answer 1

The divergence theorem allows you to turn the flux integral of your vector field $F$ through the surface $S$ into the volume integral of the divergence of $F$ over the volume $V$. In general, verifying these integrals yield the same result is a little tedious, but usually not difficult.

I will go into a lot of detail just so you see what needs to be done in general. For example, I will find parametrizations, but if you're clever you'll see that the flux through many portions of the surface will vanish.

Let's start with the flux integral. Your surface can be decomposed into four pieces: the walls alongside $y=0$ and $y=1$, the portion that lies in the $xy$-plane (which is a rectangle with $x\in[-1,1]$ and $y\in[0,1]$), and the ceiling formed by the paraboloid $z=1-x^2$ lying over that rectangle. The unit normal vector should point outward from each of these pieces. I'll consider each piece in turn; we'll need to find parametrizations:

• The wall along $y=0$. A parametrization is $\langle x, 0, z\rangle$ with $x\in[-1,1]$ and $z\in[0,1-x^2]$. The unit normal is obviously $\langle0,-1,0\rangle$. To get the integrand for the flux integral, we dot the vector field with the unit normal, which yields the opposite of the second component, $xy$. But for this part of the surface, $y=0$, so our integrand vanishes, and there is no flux through this part of the surface.
• The wall along $y=1$. A parametrization is $\langle x, 1, z\rangle$ with $x\in[-1,1]$ and $z\in[0,1-x^2]$. The unit normal is obviously $\langle0,1,0\rangle$. This time when we dot the field with the unit normal we get $-xy$, but $y=1$, so we need to integrate $$\int_{-1}^1\int_0^{1-x^2}-x\,dzdx=\int_{-1}^1-x(1-x^2)\,dx=\int_{-1}^1(x^3-x)\,dx=0$$ because $x^3-x$ is an odd function. So again there is no flux.
• The rectangle in the $xy$-plane. A parametrization is $\langle x, y, 0\rangle$ with $x\in[-1,1]$ and $y\in[0,1]$. The unit normal is obviously $\langle0,0,-1\rangle$. This time when we dot the field with the unit normal we get $-3z$, but $z=0$ on this part of the surface, so again the flux vanishes.
• The ceiling. A parametrization is $\langle x, y, 1-x^2\rangle$ with $x\in[-1,1]$ and $y\in[0,1]$. This is the only piece of the surface where the unit normal isn't obvious and needs to be calculated. To do that, we first take the cross product of the partial derivative of our parametrization with respect to $x$ and $y$. The partial with respect to $x$ is $\langle 1, 0, -2x\rangle$ and the partial with respect to $y$ is $\langle 0, 1, 0\rangle$. Crossing these gives $\langle 2x,0,1\rangle$. Dotting this with our field gives $2x(z^2-x)+3z$. But $z=1-x^2$ in our parametrization, so our integral is $$\int_{-1}^1\int_0^1\left(2x((1-x^2)^2-x)+3(1-x^2)\right)\,dxdy=\int_{-1}^1\left(2x((1-x^2)^2-x)+3(1-x^2)\right)\,dx=\frac{8}{3}$$

In summary, the flux through the surface is just the sum of the flux through the four walls, which is $0+0+0+\frac{8}{3}=\frac{8}{3}$.

Now let's do the volume integral of the divergence and check that we get the same thing. First take the divergence of your vector field; it's $$\nabla\cdot F=-1-x+3=2-x$$ Now we just want to integrate this over the volume enclosed by our surface $S$. We can parametrize the volume as $\langle x, y, z\rangle$ with $x\in[-1,1]$, $y\in[0,1]$ and $z\in[0,1-x^2]$. Then our integral is $$\int_V\nabla\cdot F dV=\int_{-1}^1\int_0^1\int_0^{1-x^2}(2-x)\,dzdydx=\int_{-1}^1\int_0^1(2-x)(1-x^2)dydx=\int_{-1}^1(2-x)(1-x^2)dx=\int_{-1}^1(2-2x^2-x+x^3)dx=\left[2x-\frac{2}{3}x^3-\frac{1}{2}x^2+\frac{1}{4}x^4\right]_{-1}^1=\frac{8}{3}$$

This agrees with our flux integral (and was much easier to do)! This illustrates a general phenomenon: integrating the divergence over the volume is often easier than doing to the painstaking flux integral through the surface.