Evaluate the triple integral of x^2 y^2 over the solid inside the paraboloid x^2 + y^2 = 2z between the planes z = 2 and z = 8.

Step-by-step solution

1. 1

   Describe the solid in cylindrical coordinates. The surface $x^2+y^2=2z$ is a paraboloid opening upward, $z=\tfrac{r^2}{2}$. "Inside the paraboloid" means $r^2\le 2z$, so slicing horizontally at height $z$ gives a disk of radius $\sqrt{2z}$:

   $2\le z\le 8,\qquad 0\le r\le \sqrt{2z},\qquad 0\le\theta\le 2\pi$

   Slicing by $z$ (rather than integrating $z$ last) is what keeps the limits simple — every cross-section is a full disk.

2. 2

   Convert the integrand and the volume element. With $x=r\cos\theta$ and $y=r\sin\theta$,

   $x^2y^2=r^4\cos^2\theta\sin^2\theta,\qquad dV=r\,dr\,d\theta\,dz$

   so the integrand picks up a total power of $r^5$:

   $\iiint_E x^2y^2\,dV=\int_2^8\!\!\int_0^{2\pi}\!\!\int_0^{\sqrt{2z}} r^5\cos^2\theta\sin^2\theta\,dr\,d\theta\,dz$

   Forgetting the Jacobian factor $r$ is the most common error and would leave $r^4$ here.

3. 3

   Separate the angular factor. Nothing in the limits depends on $\theta$, so the triple integral factors into a product of a $\theta$-integral and an $(r,z)$-integral. Using $\cos\theta\sin\theta=\tfrac12\sin 2\theta$,

   $\cos^2\theta\sin^2\theta=\frac{\sin^2 2\theta}{4}=\frac{1-\cos 4\theta}{8}$

   $\int_0^{2\pi}\cos^2\theta\sin^2\theta\,d\theta=\frac{1}{8}\left[\theta-\frac{\sin 4\theta}{4}\right]_0^{2\pi}=\frac{2\pi}{8}=\frac{\pi}{4}$

4. 4

   Do the radial integral, then the $z$-integral. For fixed $z$,

   $\int_0^{\sqrt{2z}} r^5\,dr=\frac{(2z)^3}{6}=\frac{8z^3}{6}=\frac{4z^3}{3}$

   Integrating that over $2\le z\le 8$:

   $\int_2^8 \frac{4z^3}{3}\,dz=\frac{4}{3}\left[\frac{z^4}{4}\right]_2^8=\frac{1}{3}\left(4096-16\right)=1360$

5. 5

   Multiply the two factors and sanity-check the size.

   $\iiint_E x^2y^2\,dV=1360\cdot\frac{\pi}{4}=340\pi\approx 1068.1$

   The sign is right: the integrand $x^2y^2$ is never negative, so a positive answer is mandatory. A crude bound also supports the magnitude — the solid sits inside a cylinder of radius $4$ and height $6$, where $x^2y^2\le r^4/4\le 64$, giving an upper bound of about $64\cdot\pi\cdot 16\cdot 6\approx 19300$, comfortably above $340\pi$.