Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the volume bounded by the surface $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$

I have $x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$.

Therefore, $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$ can be written as

$$(\rho^2)^2 = 2(\rho \cos\phi) [(\rho \sin\phi \cos\theta)^2 + (\rho \sin\phi \sin\theta)^2]\Rightarrow \rho = 2\cos\phi \sin^2\phi$$

so using the change of co-ordinates, I have the integral

$$\iiint \rho^2 \sin\phi d\rho d\phi d\theta$$

Now i need to find the limits of integration, but I cannot even visualize this surface. What are the limits of integration, and is my work so far correct?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

In spherical coordinates ${\varphi}\in \left[0, \; {\pi}\right].$ Because ${\rho}\geqslant{0},$ then $\cos{\varphi \geqslant{0}} \Rightarrow {{0}\leqslant {\varphi}\leqslant{\dfrac{\pi}{2}}}.$ Then limits of integration are: $$ {0}\leqslant{\theta}\leqslant{2\pi}, \\ {0} \leqslant \varphi \leqslant \dfrac{\pi}{2}, \\ {0}\leqslant{\rho}\leqslant 2\cos{\varphi} \, \sin^2{\varphi}. $$

share|improve this answer
1  
In spherical, $\varphi \in [0, \pi]$. Source: mathworld.wolfram.com/SphericalCoordinates.html –  anorton Dec 15 '12 at 0:28
    
Thanks, I edit my answer. –  M. Strochyk Dec 15 '12 at 0:29
    
Umm... the range on $\varphi$ needs to go from $0$ to $\pi$ –  anorton Dec 15 '12 at 0:36
    
@anorton From the surface equation $(x^2 + y^2 + z^2)^2 = 2z(x^2 + y^2)$ we can conclude that $z$ must be nonnegative, which is possible only if ${0}\leqslant {\varphi}\leqslant{\dfrac{\pi}{2}}$. –  M. Strochyk Dec 15 '12 at 0:46
    
Oh. Duh. Nevermind. –  anorton Dec 15 '12 at 0:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.