Question : Find the entire volume of the solid $(\frac{x}{a})^{1/2}+(\frac{y}{b})^{1/2}+(\frac{z}{c})^{1/2}=1$.
My Attempt:
After reducing it to some extent we get $V=2c\int\int_{D*} (1-(\frac{x}{a})^{1/2}-(\frac{y}{b})^{1/2})^2dxdy$. Further substituting $\frac{x}{a}=(r\cos\theta)^4, \frac{y}{a}=(r\sin\theta)^4$ (to make the use of polar-coordinates) and finding the jacobian i.e. $|J|=16abr^2(\sin\theta)^3(\cos\theta)^3$, it became $32abc\int_{0}^{2\pi}\int_{0}^{1}(1-r^2)^2(\sin\theta)^3(\cos\theta)^3r^2drd\theta$.Now we can solve the integrals separately as the limits are constants.
Here $\int_{0}^{1}(1-r^2)^2r^2dr=\frac{8}{105}$ and where I am stuck is the other left out integral. Assuming any other substitution will make it run from 1 to 1 if we assume $t=\cos\theta$ or $\sin\theta$. I think to make the use of wali's formula here I need to make it look like $\int_{0}^{\pi/2}*$.So using symmetry we have $4\int_{0}^{\pi/2}(\sin\theta)^3(\cos\theta)^3d\theta=1/3$ (using formula associated with gamma function).Then all together I got the answer $\frac{256abc}{315}$, which is incorrect. Therefore I must have done some errors in taking limits or in assuming symmetry!
The correct answer is $\frac{abc}{90}$.Thanks.
Assume $a,b,c >0$ then note your region of integration is only the first octant.
then change variables $ \frac{x}{a} =u^4$ and $\frac{y}{b} =v^4$ and $\frac{z}{c}=w^4$
then change the variables again to spherical coordinates.