# Finding the volume bounded by surface in spherical coordinates

I want to find the volume bounded by the surface given in spherical coordinates $R = 4-1\cos(\phi)$

I tried $\int_0^{2\pi} \int_0^{\pi/2} \int_0^4 (4-\cos(\phi))R^2\sin(\phi)\,dR \,d\phi\, d\theta$.

But I got the wrong answer. The volume element is given by $dV = R^2\sin(\phi)dR\,d\phi\, d\theta$. I'm assuming my limits are wrong, any ideas?

• The equation of the surface should not be in the integral, it only defines the bounds. – Kuifje Feb 27 '18 at 20:25
• Ok, so I'm integrating $R^2*sin(\phi)$ and my bounds are correct? – novo Feb 27 '18 at 20:30

$$V=\int_0^{2\pi}\int_0^{\pi}\int_0^{4-\cos\phi}R^2\sin\phi\; dRd\phi d\theta =\frac{272\pi}{3}$$
• Are you sure about the bounds, $8\pi/3$ isn't the right answer apparantly – novo Feb 27 '18 at 20:38
• There we go, I was wondering about the $1-cos(\phi)$ bound. I see you've corrected it. Thank you for the assistance, sir! – novo Feb 27 '18 at 20:42