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?
No comments:
Post a Comment