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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}

\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}}\,