Question

Find the Surface Area of the hyperbolic paraboloid f(x,y)= xy on the domain of the disk x^2 + y^2 <= 1 using polar coordinates.

Answer 1

Calc 3?

Answer 2

\[
f(x,y)= xy \to r^2\cos\theta\sin\theta \\
x^2 + y^2 <= 1 \to 0\leq r\leq 1,0\leq \theta\leq 2\pi
\]

Answer 3

\[
\Phi(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta)) = (r\cos\theta,r\sin\theta,r^2\cos\theta\sin\theta) \\
T_r\times T_\theta =\det\begin{pmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
\cos\theta &\sin\theta &2r\cos\theta\sin\theta \\
-r\sin\theta & r\cos\theta & r^2(\cos^2\theta-\sin^2\theta)
\end{pmatrix} \\
||T_r\times T_\theta || = \sqrt{(T_r\times T_\theta) \cdot (T_r\times T_\theta) }
\]\[
\iint_SdS=\iint_D||T_r\times T_\theta||drd\theta \quad D:(r,\theta)\in[0,1]\times [0,2\pi]
\]

Answer 4

The problem is I keep getting zero. \[\cos ^{2} 2\pi - \cos ^{2} 0 = 0\]

Answer 5

I'd need to see more of your work.