Question

Set up a double integral that gives the area of the surface on the graph of \(f\) over the region \(R\). $$ \begin{array}{l} f(x, y)=e^{-x} \sin y \\ R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\} \end{array} $$

Short answer

The double integral that gives the surface area of the given function over the specified region is: \[\int_{0}^{2\pi} \int_{0}^{2} \sqrt{1 + e^{-2r\cos\theta}} \, r \, dr \, d\theta.\] Note that the further computation of this integral might require numerical methods.

Step-by-step solution

Understand the Surface Area Integral Representation

The differential surface area on the surface \( z = f(x, y) \) is given by \[dA = \sqrt{1 + (f_x)^2 + (f_y)^2} \, dx \, dy\]. Here \( f_x \) and \( f_y \) are partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \) respectively.

Find Partial Derivatives

First, find the partial derivatives. For \( f(x, y) = e^{-x}\sin y \), we have \( f_x = -e^{-x}\sin y \) and \( f_y = e^{-x}\cos y \).

Setup the Surface Area Integral

Plugging these into the differential surface area under the square root gives \[dA = \sqrt{1 + \left(-e^{-x}\sin y\right)^2 + \left(e^{-x}\cos y\right)^2} \, dx \, dy = \sqrt{1 + e^{-2x}} \, dx \, dy\]

Setup Double Integral

To get the total surface area, we need to integrate over the region \( R: x^{2} + y^{2} \leq 4 \), which is a disk with radius 2 centered at origin. In polar coordinates the x and y are replaced by \( r\cos \theta \) and \( r\sin \theta \) respectively with the Jacobian \( r \). Therefore, our total surface area is given by \[\iint_{R} \sqrt{1 + e^{-2x}} \, dx \, dy = \int_{0}^{2\pi} \int_{0}^{2} \sqrt{1 + e^{-2r\cos\theta}} \, r \, dr \, d\theta\]