# MvCalc65

The given integral is $\iiint _{E}y^{2}x^{2}dV\,$

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

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:

$-1\leq y\leq 1$

$-{\sqrt {1-y^{2}}}\leq z\leq {\sqrt {1-y^{2}}}$

$0\leq x\leq 1-y^{2}-z^{2}$

Solve...

$\int _{{-1}}^{{1}}\int _{{-{\sqrt {1-y^{2}}}}}^{{{\sqrt {1-y^{2}}}}}\int _{{0}}^{{1-y^{2}-z^{2}}}y^{2}z^{2}dxdzdy\,$

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

$\int _{{-1}}^{{1}}\int _{{-{\sqrt {1-y^{2}}}}}^{{{\sqrt {1-y^{2}}}}}y^{2}z^{2}(1-y^{2}-z^{2})dzdy\,$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Factor constants and integrate term by term

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

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

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

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

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

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

${\frac {\pi }{96}}-{\frac {1}{768}}[0]\,$

Hence...$\iiint _{E}y^{2}x^{2}dV={\frac {\pi }{96}}\,$