Find the volume defined by $ 0 \leq z^2 \leq 2$ and $x^2 + 4y^2 - (2-z)^2 \leq 0$

Question

I am asked to find the volume defined by

$$ 0 \leq z^2 \leq 2\\ x^2 + 4y^2 - (2-z)^2 \leq 0 $$

How can I do that using cylindrical coordinates?

Is that minus sign before $(2-z)$ really supposed to be there? Shouldn't it be a plus sign?

Textbook answer: $\frac{4 \pi}{3}$

Thank you.

Answer 1

Should the minus sign be there. Probably. If it is.

$x^2 + 4y^2 - (2-z)^2 \le 0$ is a double cone, with an elliptical cross section.

$z^2\ge 0$ is a trivial statement.

$z^2\le 2$ is the space between 2 planes. $-\sqrt{2} \le z \le \sqrt{2}$ and $-\sqrt{2}, \sqrt{2}$ are both below the vertex of the cone.

we have a fustrum.

lower base:

$x^2 + 4y^2 = (2+\sqrt 2)^2\\ A_l = \frac 12 \pi (2+\sqrt 2)^2$

upper base:

$x^2 + 4y^2 = (2-\sqrt 2)^2 A_u = \frac 12 \pi (2-\sqrt 2)^2$

$V =$$ \frac 13 (A_l (2+\sqrt 2) - A_u(2-\sqrt 2))\\ \frac 16 \pi [(2+\sqrt2)^3 - (2-\sqrt 2)^3)$

which is greater than $\frac {4 \pi}{3}$ (it is closer to $\frac {20 \pi}{3}$)

If you want to use calculus.

$x = r cos t\\ y = \frac 12 r cos t\\ z = z\\ dz\,dy\,dx = \frac 12 r \,dz\,dr\,dt$

$x^2 + 4y^2 - (2-z)^2 \le 0$ becomes

$r^2 \le (2-z)^2\\ r \le (2-z)$

$\int_0^{2\pi}\int_{-\sqrt{2}}^{\sqrt{2}}\int_0^{2-z} \frac 12 r \,dr\,dz\,dt\\ \int_0^{2\pi}\int_{-\sqrt{2}}^{\sqrt{2}} \frac 14 (2-z)^2 \,dz\,dt\\ \int_0^{2\pi} -\frac 1{12} (2-z)^3\,dt\\ \frac 1{12} \int_0^{2\pi} (2+\sqrt{2})^3-(2-\sqrt{2})^3\,dt\\ \frac {\pi}{6} [(2+\sqrt{2})^3-(2-\sqrt{2})^3]$

$\frac{14}{3} \pi\sqrt{2}$

Answer 2

Let's integrate this in terms of $z$. Let $A(z)$ be the area of the horizontal cross section of $x^2+4y^2-(2-z)^2\leq 0$ at height $z$. Instead of integrating from $-\sqrt{2}\leq z\leq\sqrt{2}$, we will integrate from $4-\sqrt{2}\leq z \leq 4+\sqrt{2}$. (This is only so that we do not have to deal with ugly negative signs. Since the graph $x^2+4y^2-(2-z)^2= 0$ is symmetric across the plane $z=2$, the answer will be the same.)

Then we are trying to find $\displaystyle\int_{4-\sqrt{2}}^{4+\sqrt{2}}A(x)\text{ }dz$.

Since $x^2+4y^2-(2-z)^2= 0$ is an elliptical paraboloid, each horizontal cross section is an ellipse:

$$x^2+4y^2-(2-z)^2= 0$$ $$x^2+4y^2=(2-z)^2$$ $$\left(\frac{x}{2-z}\right)^2+\left(\frac{2y}{(2-z)}\right)^2=1$$

Ergo, $A(z)=\displaystyle\frac{\pi(2-z)^2}{2}$

Finally, $\displaystyle\int_{4-\sqrt{2}}^{4+\sqrt{2}}\frac{(2-z)^2\pi}{2}\text{ }dz=\frac{\pi\left(4+\sqrt{2}-2\right)^3}{6}-\frac{\pi\left(4-\sqrt{2}-2\right)^3}{6}=\frac{14\sqrt{2}\pi}{3}$