Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{array}{l} f(x, y)=2+\frac{2}{3} y^{3 / 2} \\ R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 2-x\\} \end{array} $$

Short answer

The surface area of the given surface over the given region is \(\frac{32}{15}\) square units.

Step-by-step solution

Understand the Formulas to Use

The main formula to use in this context is \(S=\int \int_{R} d s\), where \(ds = \sqrt{1+ (\partial f/\partial x)^{2}+(\partial f/\partial y)^{2}} \, dx \,dy \). So first, we have to compute the partial derivatives of \(f(x, y)\).

Compute the Partial Derivatives

Starting with \(f(x, y)=2+\frac{2}{3} y^{3 / 2}\), we can compute the partial derivatives. We see that \(f_x=\frac{df}{dx}=0\) since there is no x term in the function. For the derivative in terms of y, we get \(f_y=\frac{df}{dy}=y^{1/2}\).

Substitute into the Formula for ds

After calculating the partial derivatives, we substitute them into the formula for ds. So \(ds= \sqrt{ 1+ \left(0\right)^2+\left(y^{1/2}\right)^2 } = \sqrt{1+y}\).

Set Up the Integral for Surface Area

Next, we substitute this into the surface area integral, and we get \(S=\int \int_{R} \sqrt{1+y} \, dx \,dy\).

Define the Limits of Integration According to Region R

The region \(R=\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 2-x\}\) specifies the limits of integration. So we set up the limits: \(S = \int_{0}^{2} \int_{0}^{2-x} \sqrt{1+y} \, dy \,dx\).

Compute the Integral

At the end, we computed the surface integral to find the surface area. Simultaneously compute the integral from inside to outside: first in terms of \(y\), and then in terms of \(x\). After computing, we get the value of \(S=\frac{32}{15}\).