Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) \(f(x, y)=2 x+2 y\) \(R:\) triangle with vertices (0,0),(2,0),(0,2)

Short answer

The area of the surface given by \( z = f(x, y) = 2x + 2y \) over the specified triangular region is 4 square units.

Step-by-step solution

Compute the partial derivatives

First, compute the partial derivatives of \( f(x, y) \) with respect to x and y respectively. For our given function \( f(x, y) = 2x + 2y \), we have \( f_x = 2 \) and \( f_y = 2 \).

Substitute in the formula

Substitute the values of the partial derivatives in the formula for the surface integral: \(\int \int_R \sqrt{1 + (f_x)^2 + (f_y)^2} dA = \int \int_R \sqrt{1 + 2^2 + 2^2} dA \). After simplifying we get, \(\int \int_R \sqrt{9} dA = \int \int_R 3 dA.\)

Calculate the surface area

Finally, calculate the surface area by integrating the function over the region R. The triangle is in the xy-plane with vertices at (0,0), (2,0), and (0,2). Thus, our limits of integration for x will be from 0 to 2, and for y from 0 to (2-x). Therefore, performing the double integral gives us: \(\int \int_R 3 dA = \int_{0}^{2} \int_{0}^{2-x} 3 dy dx = 3 \int_{0}^{2} (2-x) dx = 3 \cdot [2x - 1/2 x^2]_0^2 = 6 - 2 = 4.\)