# MvCalc65

The given integral is $\displaystyle \iiint_E y^2 x^2 dV\,$

Where E is the region bounded by the paraboloid $\displaystyle x= 1 - y^2 - z^2$ and the plane $\displaystyle x = 0$

First, we'll set up the limits of integration. The bounds are a paraboloid along the x-axis intersected by the y-z plane. So, the limits are:

$\displaystyle -1 \le y \le 1$

$\displaystyle -\sqrt{1-y^2} \le z \le \sqrt{1-y^2}$

$\displaystyle 0 \le x \le 1 - y^2 - z^2$

Solve...

$\displaystyle \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{0}^{1-y^2-z^2}y^2 z^2 dxdzdy \,$

$\displaystyle \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} y^2z^2[x]_{0}^{1-y^2-z^2} dzdy \,$

$\displaystyle \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} y^2z^2(1-y^2-z^2)dzdy \,$

Now changing to polar cordinates, on the y-z plane: $\displaystyle y=r\cos\theta , z=r\sin\theta , -y^2-z^2=-r^2 \,$

and for the limits of integration r = 1 and the intersection is the unit circle on the y-z plane so:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1} r^2\cos^2\theta{r^2}\sin^2\theta(1-r^2)rdrd\theta \,$

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1} \cos^2\theta\sin^2\theta(r^5-r^7)drd\theta \,$

$\displaystyle \int_{0}^{2\pi} \cos^2\theta\sin^2\theta d\theta\int_{0}^{1} (r^5-r^7) dr \,$

$\displaystyle \int_{0}^{2\pi} \cos^2\theta\sin^2\theta d\theta[[\frac{1}{6}r^6]-[\frac{1}{8}r^8]]_{0}^{1}\,$

$\displaystyle \int_{0}^{2\pi} \cos^2\theta\sin^2\theta d\theta(\frac{1}{6}-\frac{1}{8})\,$

$\displaystyle \frac{1}{24}\int_{0}^{2\pi} \cos^2\theta\sin^2\theta d\theta\,$

NOTE THAT: $\displaystyle \sin^2\theta\cos^2\theta=(\sin\theta\cos\theta)^2\,$ and $\displaystyle \sin(2\theta)=2\sin\theta\cos\theta \,$ by double-angle formula. Divide by 2 and square both sides

So that... $\displaystyle \sin^2\theta\cos^2\theta= \frac{1}{4}\sin^2(2\theta) \,$ Plug this into integral.

$\displaystyle \frac{1}{24}\int_{0}^{2\pi}\frac{1}{4}\sin^2(2\theta) d\theta\,$

$\displaystyle \frac{1}{96}\int_{0}^{2\pi}\sin^2(2\theta) d\theta\,$

Substitute u=2θ so 1/2du=dθ and when θ = 2π then u = 4π and when θ = 0 then u = 0

$\displaystyle \frac{1}{192}\int_{0}^{4\pi}\sin^2(u)du\,$

$\displaystyle \frac{1}{192}\int_{0}^{4\pi}(\frac{1}{2}-\frac{1}{2}\cos(2u))du\,$ By half-angle formula

Factor constants and integrate term by term

$\displaystyle \frac{1}{192}\int_{0}^{4\pi}\frac{1}{2}du-\frac{1}{384}\int_{0}^{4\pi}\cos(2u)du \,$

$\displaystyle \frac{1}{384}[u]_{0}^{4\pi}-\frac{1}{384}\int_{0}^{4\pi}\cos(2u)du \,$

$\displaystyle \frac{\pi}{96}-\frac{1}{384}\int_{0}^{4\pi}\cos(2u)du \,$

Substitute v=2u so 1/2dv=du and when u = 4π then v = 8π and when u = 0 then v = 0

$\displaystyle \frac{\pi}{96}-\frac{1}{768}\int_{0}^{8\pi}\cos(v)dv \,$

$\displaystyle \frac{\pi}{96}-\frac{1}{768}[\sin(v)]_{0}^{8\pi} \,$

$\displaystyle \frac{\pi}{96}-\frac{1}{768} \,$

Hence...$\displaystyle \iiint_E y^2 x^2 dV = \frac{\pi}{96} \,$