Evaluate the triple integral of x over the solid bounded by the cone x^2 = y^2 + z^2 and the paraboloid x = y^2 + z^2.

Step-by-step solution

1. 1

   Exploit the symmetry: both surfaces depend on $y$ and $z$ only through $y^2+z^2$. Set

   $r^2=y^2+z^2$

   Then the cone becomes $x^2=r^2$ and the paraboloid becomes $x=r^2$. The solid is a surface of revolution about the $x$-axis, so the natural coordinates are cylindrical coordinates wrapped around that axis, not the usual $z$-axis.

2. 2

   Set up the rotated cylindrical coordinates. Take

   $y=r\cos\theta,\qquad z=r\sin\theta,\qquad x=x$

   with $r\ge 0$ and $0\le\theta\le 2\pi$. The Jacobian is the same as always, so

   $dV=r\,dx\,dr\,d\theta$

   Forgetting that extra factor of $r$ is the classic mistake here.

3. 3

   Find where the two surfaces meet. The enclosed region sits in $x\ge 0$, where the cone is $x=r$. Equating the two:

   $r=r^2\qquad\Longrightarrow\qquad r(r-1)=0\qquad\Longrightarrow\qquad r=0\ \text{ or }\ r=1$

   So the solid lives over the disc $0\le r\le 1$ in the $yz$-plane, and the curves meet at the origin and at $x=1$.

4. 4

   Decide which surface is the lower boundary. For $0<r<1$ we have $r^2<r$, so the paraboloid $x=r^2$ lies to the left of the cone $x=r$. For a fixed $(r,\theta)$ the variable $x$ therefore runs from $r^2$ up to $r$:

   $\iiint_E x\,dV=\int_0^{2\pi}\!\!\int_0^1\!\!\int_{r^2}^{r} x\,r\,dx\,dr\,d\theta$

5. 5

   Integrate in $x$, then in $r$. The inner integral is

   $\int_{r^2}^{r}x\,r\,dx=r\left[\frac{x^2}{2}\right]_{r^2}^{r}=\frac{r^3-r^5}{2}$

   and then

   $\int_0^1\frac{r^3-r^5}{2}\,dr=\frac{1}{2}\left[\frac{r^4}{4}-\frac{r^6}{6}\right]_0^1=\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}$

6. 6

   Finish with the angular integral. The integrand does not involve $\theta$, so the last integration just multiplies by the full turn:

   $\int_0^{2\pi}\frac{1}{24}\,d\theta=\frac{2\pi}{24}=\frac{\pi}{12}$

   $\boxed{\iiint_E x\,dV=\frac{\pi}{12}\approx 0.2618}$

   A numerical Riemann sum over the same region gives $0.26180$, confirming the result. Since $x\ge 0$ throughout the solid, a positive answer was expected.