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

   Recognise the integrand as $r^2$. The combination $x^2+y^2$ is exactly the squared cylindrical radius, so cylindrical coordinates are the natural choice — the integrand becomes a single power of $r$ with no angular dependence at all:

   $x^2+y^2=r^2,\qquad dV=r\,dr\,d\theta\,dz$

   This quantity is the integrand behind the moment of inertia of the solid about the $z$-axis, which is why it appears so often.

2. 2

   Set up the limits by horizontal slices. Inside the paraboloid $z=\tfrac{r^2}{2}$ means $r\le\sqrt{2z}$, and the planes cap $z$ between $2$ and $8$:

   $\iiint_E\left(x^2+y^2\right)dV=\int_2^8\!\!\int_0^{2\pi}\!\!\int_0^{\sqrt{2z}} r^2\cdot r\,dr\,d\theta\,dz=\int_2^8\!\!\int_0^{2\pi}\!\!\int_0^{\sqrt{2z}} r^3\,dr\,d\theta\,dz$

3. 3

   Integrate radially. For a fixed height $z$,

   $\int_0^{\sqrt{2z}} r^3\,dr=\left[\frac{r^4}{4}\right]_0^{\sqrt{2z}}=\frac{(2z)^2}{4}=z^2$

   The $(2z)^2/4$ collapsing to exactly $z^2$ is a small piece of luck built into the coefficient $2$ in $x^2+y^2=2z$, and it makes the rest of the problem trivial.

4. 4

   Sweep the angle and then the height. Since nothing depends on $\theta$, that integral contributes a factor $2\pi$:

   $\int_2^8\!\!\int_0^{2\pi} z^2\,d\theta\,dz=2\pi\int_2^8 z^2\,dz=2\pi\left[\frac{z^3}{3}\right]_2^8=2\pi\cdot\frac{512-8}{3}=2\pi\cdot 168$

5. 5

   State the answer and check it against the volume.

   $\iiint_E\left(x^2+y^2\right)dV=336\pi\approx 1055.6$

   A quick plausibility check: the volume of the same solid is $\int_2^8 \pi(2z)\,dz=\pi(64-4)=60\pi$, so the average value of $x^2+y^2$ over the solid is $\tfrac{336\pi}{60\pi}=5.6$. That is a believable mean squared radius for a region whose radius ranges from $2$ to $4$.