Calculating the volume between surfaces by integration

Question

Calculate the volume between these surfaces:

• $x+y+z=1/2$

• $x+y-z=-1/2$

• $z=x^2+y^2$

I'd really appreciate a solution that shows steps and explains the thinking process between them.

Answer 1

First of all we observe that the first two surfaces are planes. And they are symmetric around the line $x + y = 0$. We can focus on one of these planes (the one that is highest), and simply ignore the other. My choice is to take $y > -x$.

Secondly we need a strategy to calculate the volume. In this case the obvious choice is to take the difference in $z$ values between two surfaces and to integrate over the variables $x$ and $y$.

In order to perform the integration, we must find the integration bounds. These can be derived from the fact that we integrate until the surfaces intersect, i.e. where the difference in $z$ value equals zero. We have $z = x+y + 1/2 = x^2 + y^2$, which means that the bounds are given by $(x-1/2)^2 + (y-1/2)^2 = 1$. This is the equation for a circle with radius $1$ and center $(1/2, 1/2)$. However, since we work in the half-space $y > -x$, part of this circle is chopped off.

My choice is to divide the truncated circle into two sections: (a) three quarters of the circle described above, and (b) a triangle between the points $(-1/2, 1/2)$, $(1/2, 1/2)$ and $(1/2, -1/2)$.

For the first one we introduce polar coordinates $(r, \phi)$ around the point $(1/2, 1/2)$:

$$I1 = \int_{0}^{1}dr \space r \int_{-\pi /2}^{\pi}d\phi ^\space (x+y+ 1/2 - x^2-y^2)$$

Substituting $x = 1/2 + r\cos(\phi)$ and $y = 1/2 + r\sin(\phi)$ the integral simplifies to

$$I1 = \int_{0}^{1}dr \int_{-\pi /2}^{\pi}d\phi ^\space r(1-r^2) = \frac{3}{8} \pi$$

$$I2 = \int_{-1/2}^{1/2}dx \int_{-x}^{1/2} dy \space (x+y+ 1/2 - x^2-y^2) = 1/3$$

The net volume in region $y > -x$ is equal to $\frac{3}{8} \pi + 1/3$. Due to symmetry the same value is found in the region $y < -x$. Hence the total volume is $\frac{3}{4}\pi + 2/3$.