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I want to calculate the volume enclosed by $z^2 = 1+x^2+y^2$ and the plane $z=2$.

When $z = 2$, $x^2+y^2 = 3 \rightarrow r = \sqrt{3}$ So I have set up the integral in polar coordinates as:

$$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r\sqrt{1+r^2}drd\theta$$

Solving this integral I get $14\pi/3$ however this gives us the area under the surface and we want the area above the surface so I minus the area of the cylinder height $2$ and radius $ \sqrt{3}$ i.e. $A = \pi r^2h = 6\pi$ therefore the final answer should be $8\pi/3$ but according to the answer I am 2 times the correct answer? Where did I go wrong?

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  • $\begingroup$ Just integrate the difference between the upper and lower function: $\sqrt{1+r^2}-2$. $\endgroup$ – orion Jun 4 '15 at 11:49
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For any $z\in[1,2]$, the section is a circle with square radius equal to $z^2-1$, hence: $$ V = \pi\int_{1}^{2}(z^2-1)\,dz = \frac{4\pi}{3}$$ by Cavalieri's principle.

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