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Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.

I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $\theta$ I got $\tan^{-1}4\leq\theta\leq\pi-\tan^{-1}4$.

$$\begin{align}\iiint_E\sqrt{3x^2+3z^2}\,dV &=\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}\int_8^{2r^2}\sqrt{3}\cdot r\ dy \ r\ dr\ d\theta\\&=2\sqrt3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}r^2(r^2-4)dr\ d\theta\\&=2\sqrt3\ 8^3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\csc^3\theta\bigg(\dfrac{8^2\csc^2\theta}{5}-\dfrac{4}{3}\bigg)d\theta\end{align}$$

Now, this is ugly to get through, please help, is there any efficient way to carry out this problem?

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  • $\begingroup$ how about Weirstrass Substitution? $\endgroup$
    – dfnu
    Commented Oct 3, 2019 at 16:29
  • $\begingroup$ @dfnu I've not heard of this, any weblink would help. $\endgroup$
    – mnulb
    Commented Oct 3, 2019 at 16:30
  • $\begingroup$ en.m.wikipedia.org/wiki/Tangent_half-angle_substitution $\endgroup$
    – dfnu
    Commented Oct 3, 2019 at 16:45
  • $\begingroup$ All your integration bounds are way off... You should have $2r^2 \le y \le 8$ to begin with, since the plane is above the paraboloid, if $y$ is “up”. And the solid is rotationally symmetric about the $y$ axis, so the angle $\theta$ goes a whole lap, from $0$ to $2\pi$ (say). And the radial variable $r$ is the distance from $(0,0)$ to $(x,z)$ in the $xz$ plane, not the distance from $(0,0,0)$ to $(x,y,z)$ in $xyz$ space, so it's just $0 \le r \le 2$. $\endgroup$ Commented Oct 3, 2019 at 20:00

2 Answers 2

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Observe that the solid is symmetric with respect to the y-axis. So, it is efficient to integrate using cylindrical coordinates along the y-direction. Rewrite the original integral as

$$\iiint_E\sqrt{3x^2+3z^2}\,dV =\int_0^8 dy \int_0^{2\pi}d\theta\int_0^{\sqrt{y/2}} \sqrt{3}r\>rdr$$ $$=2\pi\sqrt 3 \int_0^8 dy \frac13 \left(\frac y2 \right)^{3/2}= \frac{256\sqrt 3}{15}\pi $$

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We do not have to convert to other coordinate system. Since for a fixed value of y, the cross section is a circle with radius of $\sqrt{\frac{y}{2}}$. Then the integral becomes

$$\int_{y=0}^{8} \int_{x=-\sqrt{\frac{y}{2}}}^{\sqrt{\frac{y}{2}}} \int_{z=-\sqrt{\frac{y}{2}-x^2}}^{\sqrt{\frac{y}{2}-x^2}} dz dx dy$$ $$= \int_{y=0}^{8} \frac{\pi}{2} \frac{\sqrt{3}}{\sqrt{2}} y^{\frac{3}{2}} dy$$ $$= \frac{128\sqrt{3}\pi}{5}$$

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