Find volume of solid bounded by given surfaces. $z=a+x,z=-a-x,x^2+y^2=a^2$

Question

Find volume of solid bounded by given surfaces.

$$z=a+x, \qquad z=-a-x, \qquad x^2+y^2=a^2$$

This is the solid. We can find volume of solid that has positive $z$ value and multiply by $2$. And for finding volume of that solid I thought doing double integral where domain is circle and function is that ellipse. $\int_{-a}^a dy \int_{-a}^a f(x,y)dx$. Where $f(x,y)$ is intersection of plane $z=a+x$ and $x^2+y^2=a^2$. Is my thinking right? When I want to find intersection I substitute $x=z-a$ into $x^2+y^2=a^2$ I get $z=\frac{a+\sqrt{a^2-y^2}}{2}$ which when I graph does not look like ellipse. Can you help?

Answer 1

In cylindrical coordinates, the cutting plane for $z>0$ is $z=a+r\cos \theta$. The volume is then $$V =2 \int_0^{2\pi} \int_0^a z\ rdr d\theta =2\pi a^3 $$ Note that, geometrically, the two symmetric parts above/below the $xy$-plane configures to a regular cylinder of cross-section area $\pi a^2 $ and height $2a$, hence the volume $2\pi a^3$.

Answer 2

Intersection of planes is a line given by, $z = a + x = - a - x \implies x = - a, z = 0$
At intersection with the cylinder, $(-a)^2 + y^2 = a^2 \implies y = 0$

So, both planes and the cylinder intersect at a single point $(-a, 0, 0)$. That means the projection of the region in xy-plane is circle $x^2 + y^2 \leq a^2, z = 0$

The integral to find volume is given by,

$ \displaystyle V = \iint_{x^2 + y^2 \leq a^2} 2 |a + x| ~ dx ~ dy$

Note that if $a \lt 0, a + x $ is always negative (or zero) over the region and if $a \gt 0, a + x$ is always non-negative over the region.

Now as $x$ is an odd function, $[f(-x) = - f(x)]$, and the region is symmetric to yz plane, its integral over $x \gt 0$ will cancel out the integral over $x \lt 0$.

So, $ \displaystyle V = 2 |a| \iint_{x^2 + y^2 \leq a^2} dx ~ dy = 2 \pi |a|^3$