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?

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

$$ V=\int_0^{2\pi}\int_0^{\pi}\int_0^{4-\cos\phi}R^2\sin\phi\; dRd\phi d\theta =\frac{272\pi}{3} $$

  • $\begingroup$ Are you sure about the bounds, $8\pi/3$ isn't the right answer apparantly $\endgroup$ – novo Feb 27 '18 at 20:38
  • $\begingroup$ There we go, I was wondering about the $1-cos(\phi)$ bound. I see you've corrected it. Thank you for the assistance, sir! $\endgroup$ – novo Feb 27 '18 at 20:42

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