Evaluate x2 dv, e where e is the solid that lies within the cylinder x2 + y2 = 4, above the plane z = 0, and below the cone z2 = 9x2 + 9y2.

Convert to cylindrical coordinates, setting

[tex]\begin{cases}x=r\cos t\\y=r\sin t\\z=z\end{cases][/tex]

Then

[tex]\displaystyle\iiint_Ex^2\,\mathrm dV=\int_{t=0}^{t=2\pi}\int_{r=0}^{r=2}\int_{z=0}^{z=3r}r^3\cos^2t\,\mathrm dz\,\mathrm dr\,\mathrm dt=\frac{96\pi}5[/tex]

Answer Link

codiepienagoya codiepienagoya · Answer 2 · 2021-11-21T18:38:50+01:00

Following are the solution to the given equation:

Given

[tex]\int \int \int x^2 dV[/tex] wherein E is indeed the solid that exists inside the cylinders [tex]x^2+y^2=4[/tex] above surface [tex]z = 0[/tex] , that is below the cones [tex]z^2 = 9x^2+9y^2[/tex] E's precession through into xy-plane is [tex]x^2+y'^2 \leq 4 \to r \leq 2[/tex] with [tex]\theta[/tex] in[tex][0,2 \pi ] \\\\[/tex]

[tex]\int \int \int x^2 \ dV =\int^{2\pi}_{0} \int^{2}_{0}\int^{3r}_{z= 0} r^2 \cos^2 \theta r \ dz \ dr \ d\theta \\\\[/tex]

[tex]=\int^{2\pi}_{0} \int^{2}_{0}\int^{3r}_{z= 0} r^3 \cos^2 \theta \ dz \ dr \ d\theta \\\\= \int^{2\pi}_{0} \int^{2}_{0} 3r^4 \cos^2 \theta \ dr \ d\theta \\\\= 3 \int^{2\pi}_{0} \frac{1}{2} \frac{2^5}{5}(1+\cos\ 2 \theta) \ d\theta \\\\= \frac{48}{5}[ (\theta +\frac{1}{2}\sin 2 \theta) ]^{2\pi}_{0} \\\\= \frac{96\pi}{5}[/tex]

Therefore, the final answer is "[tex]\frac{96\pi}{5}[/tex]" .

Learn more:

brainly.com/question/6269101