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A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates. Doing so I got $p^2=sin(\phi)cos(\theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?

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  • $\begingroup$ Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute. $\endgroup$ – Aaron Dec 11 '18 at 1:48
  • $\begingroup$ Can you explain how that is? $\endgroup$ – Skrrrrrtttt Dec 11 '18 at 2:10
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    $\begingroup$ Does this answer your question? Volume enclosed by $(x^2+y^2+z^2)^2=x$ $\endgroup$ – J.-E. Pin Jan 28 at 8:44
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You can use polar coordinates to evaluate the integral. However, you should swap the roles of $x,y,z$ directions to simplify the expressions.

In polar coordinates $(r,\theta,\phi) \mapsto ( r\cos\theta, r\sin\theta\cos\phi, r\sin\theta\sin\phi)$ where $r \ge 0$, $\theta \in [0,\pi]$ and $\phi \in [0,2\pi)$. The surface becomes $$(x^2+y^2+z^2)^2 = x \iff r^4 = r\cos\theta$$ Since $r \ge 0$, the relevant range of $\theta$ is $[0,\pi/2]$ where $\cos\theta \ge 0$. The volume we seek becomes

$$\int_0^{2\pi} \int_0^{\pi/2} \left(\int_0^{\sqrt[3]{\cos\theta}} r^2 dr\right) \sin\theta d\theta d\phi =\int_0^{2\pi}\int_0^{\pi/2} \left[ \frac13 r^3 \right]_{r=0}^{\sqrt[3]{\cos\theta}} \sin\theta d\theta d\phi\\ = \frac{2\pi}{3}\int_0^{\pi/2} \cos\theta\sin\theta d\theta = \frac{\pi}{3} $$

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For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get $$r^2=\sqrt x - x^2$$ There’s no need for a plus or minus because everything is squared and therefore positive. Using the classic volume of a solid of revolution technique, we get $$\pi \int_0^1 r^2 dx = \int_0^1 \sqrt x - x^2 dx$$ (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)

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Try $\int_0^1 \pi r(x)^2\,dx$ where $r(x)$ is obtained from @Aaron's hint.

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