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Find the area of the portion of the surface $z=x^2-y^2$ in $\mathbb{R}^3$ which lies inside the solid cylinder $x^2+y^2\le1$.

I parametrized the surface as $x=r\cos\theta$,$y=r\sin\theta$,$z=r^2\cos2\theta$.Then

$$\phi(r,\theta)=(r\cos\theta, r\sin\theta, r^2\cos2\theta) $$

then $$|(\phi_r \times \phi_\theta)|=r\sqrt{4r^2+1}$$ .

Then I am not sure where the limits should vary.

Thomas Russell
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tattwamasi amrutam
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1 Answers1

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You've chosen to parametrize using polar coordinates, which is a good idea.

The solid cylinder $x^2 + y^2 \leq 1$ extends above the disk $D$ defined by $x^2 + y^2 \leq 1$, which is located in the $xy$-plane. For each point $(x,y)$ in $D$, you have a corresponding point $(x,y,z) = (x,y,x^2 - y^2)$ directly above or below it which is on the surface.

So really, you just need to answer the following question: How can $(r,\theta)$ be chosen so as to cover the entire disk $D$ exactly once? Therefore you need to allow $\theta$ to vary from $0$ to $2\pi$ (for example), and $r$ to vary from $0$ to $1$.

user121926
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