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I am trying to solve the following problem:

Compute the volume of the solid bounded by the $ \ z \ = \ 3 \sqrt{x^2 + y^2} \ $ , the plane $ \ z = 0\ $ , and the cylinder $ \ x^2 +(y−1)^2 \ = \ 1 \ $ .

I know I need to use a triple integral, integrating over $ \ z, r, $ and $ \ \theta \ $, but I am not sure how to account for the $ \ y \ $ shift. How do I do this?

colormegone
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user145277
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  • Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $\theta$. – David H May 04 '14 at 03:05
  • Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r. – user145277 May 04 '14 at 16:24

2 Answers2

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The problem is to compute the volume of the region $R$ bounded by the three surfaces,

$$\begin{cases}z=0,\\z=3\sqrt{x^2+y^2},\\x^2+(y-1)^2=1.\end{cases}$$

This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(\rho\cos{\phi},\rho\sin{\phi},z)$, the three surface equations become

$$\begin{cases}z=0,\\z=3\sqrt{(\rho\cos{\phi})^2+(\rho\sin{\phi})^2},\\(\rho\cos{\phi})^2+(\rho\sin{\phi}-1)^2=1,\end{cases}$$

or

$$\begin{cases}z=0,\\z=3\rho,~~~0\leq\rho\\\rho=2\sin{\phi},~~~0\leq\phi<\pi.\end{cases}$$

Thus, the volume of $R$ can be found by evaluating the following triple integral:

$$V=\iiint_{R}dV\\ =\int_{0}^{\pi}d\phi\int_{0}^{2\sin{\phi}}d\rho\,\rho\int_{0}^{3\rho}dz\\ =\int_{0}^{\pi}d\phi\int_{0}^{2\sin{\phi}}d\rho\,(3\rho^2)\\ =\int_{0}^{\pi}d\phi\,(2\sin{\phi})^3\\ =\frac{32}{3}.$$

David H
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HINT

Convert to cylindrical/polar coordinates

$$ z=3 \,r; z = 6 \sin \theta; \, r = 2 \sin \theta, $$

the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).

Narasimham
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