# volume of surface given by $(x^2+y^2+z^2)^2=x$

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?

• 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. – Aaron Dec 11 '18 at 1:48
• Can you explain how that is? – Skrrrrrtttt Dec 11 '18 at 2:10
• Does this answer your question? Volume enclosed by $(x^2+y^2+z^2)^2=x$ – J.-E. Pin Jan 28 at 8:44

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
$$\int_0^{2\pi} \int_0^{\pi/2} \left(\int_0^{\sqrt{\cos\theta}} r^2 dr\right) \sin\theta d\theta d\phi =\int_0^{2\pi}\int_0^{\pi/2} \left[ \frac13 r^3 \right]_{r=0}^{\sqrt{\cos\theta}} \sin\theta d\theta d\phi\\ = \frac{2\pi}{3}\int_0^{\pi/2} \cos\theta\sin\theta d\theta = \frac{\pi}{3}$$
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 $$x$$)
Try $$\int_0^1 \pi r(x)^2\,dx$$ where $$r(x)$$ is obtained from @Aaron's hint.