Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{array}{l} f(x, y)=\ln |\sec x| \\ R=\left\\{(x, y): 0 \leq x \leq \frac{\pi}{4}, 0 \leq y \leq \tan x\right\\} \end{array} $$

Short answer

The surface area of the given region is \(\frac{1}{2} (\sqrt{2}-\ln(\sqrt{2}+1))\) square units.

Step-by-step solution

Compute Partial Derivatives

The first step is to calculate the partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\). In this case the function is \(f(x, y) = \ln |\sec(x)|\). The partial derivative with respect to \(y\) is 0, because \(f(x, y)\) doesn't depend on \(y\), and that with respect to \(x\) is \(f_x=\frac{\tan x}{\sec x}=\sin x\).

Set up the Integral for the Surface Area

The formula for the surface area of a region \(S\) given by \(z=f(x, y)\) is \(A(S)= \int_S \sqrt{1 + (f_x)^2 + (f_y)^2} dA\). For this function, the formula gets simplified to \(\int_R \sqrt{1 + (\sin x)^2} dx dy\). The region \(R\) is given as \(0 \leq x \leq \frac{\pi}{4}, 0 \leq y \leq \tan x\). The integral for the surface area thus becomes \(\int_0^{\frac{\pi}{4}} \int_0^{\tan x} \sqrt{1 + (\sin x)^2} dy dx\).

Evaluate the Integral

Now perform the integral: \(\int_0^{\frac{\pi}{4}} \sqrt{1 + sin^2(x)} \cdot \tan(x) dx\). This integral is not trivial to compute, but using common integral properties and a substitution like \(u = \tan(x)\), it can be shown that the integral equals \(\frac{1}{2} (\sqrt{2}-\ln(\sqrt{2}+1))\).