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The following question is from a recent multivariable calculus exam from my university:

Find the volume of the solid enclosed by the following surfaces: $$x^2+y^2=z^2, (x-1)^2+y^2=1, x^2+y^2=4,$$ and $z=xy+10$.

The first thing that I notice is that the cylinder $x^2+y^2=4$ is redundant because it contains the cylinder $(x-1)^2+y^2=1$, but when I try to reduce the triple integral to a double by looking at a fixed $z=t$ axis ($t$ fixed), I get that the smallest $t$ such that the intersection between the circle $x^2+y^2=t^2$ and hyperbola $y=\frac{t}{x}$ is non-empty is between $3$ and $4$ (specifically, the root of the equation $t^4-4(10-t)^2=0$, which is very difficult to solve exactly), but this doesn't make any sense since the circle $x^2+y^2=t^2$ then contains both of the smaller circles, and the hyperbola will never intersect either of the "smaller" circles. I've also tried fixing $x$ and $y$, but to no avail.

Is there a different, better approach to this problem? Maybe a coordinate substitution? I'd appreciate any help or hints.

1 Answers 1

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The projection of the solid in the $xy$ plane is the following region $D$:enter image description here

$D$ can be described as the union of $D_1$ and $D_2$, where $D_1$ is the part of the $D$ on the left of $x=0$ and $D_2$ the part on the right of $x=0$: $$ D_1=\{(r,\theta)\;|\;\pi/2 \le \theta \le 3\pi/2, \; 0\le r \le 2 \}\\ D_2=\{(r,\theta)\;|\;-\pi/2 \le \theta \le \pi/2, \; 2\cos\theta \le r \le 2 \}\\ $$

It follows that the volume equals $$ V = \iint_{D_1} xy+10-\sqrt{x^2+y^2}\; dA+\iint_{D_2} xy+10-\sqrt{x^2+y^2}\; dA $$ Switching to polar coordinates yields $$ V=\int_{\pi/2}^{3\pi/2}\int_{0}^{2}(r^2\cos\theta\sin\theta+10-r)r\; dr d\theta+\int_{-\pi/2}^{\pi/2}\int_{2\cos\theta}^{2}(r^2\cos\theta\sin\theta+10-r)r\; dr d\theta = \frac{32}{9}+\frac{74\pi}{3} \approx 81.048 $$

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    How do you know $z=xy+10$ is above the -cone-. – 2017-02-02
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    Yes that is a good point, it needs to be verified (its not a sphere but a cone). The big cylinder and the cone intersect at $z=2$. Clearly, the cone is the inferior bound. – 2017-02-02
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    In the description of $D_{2}$, why is it $\cos{\theta}\leq r \leq 2$? I get $(x-1)^2+y^2 \geq 1 \implies r^2\cos^2{\theta}-2r\cos{\theta}+1+r^2\sin^2{\theta} \geq 1 \implies r \geq 2\cos{\theta}$. – 2017-02-03
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    You are correct! Thanks for pointing that out, I have edited! – 2017-02-03