Question

Determine which value best approximates the surface area of \(z=f(x, y)\) over the region \(R .\) \(f(x, y)=\frac{1}{4} \sqrt{x^{2}+y^{2}}\) \(R:\) circle bounded by \(x^{2}+y^{2}=9\) (a) -100 (b) 150 (c) \(9 \pi\) (d) 55 (e) 500

Short answer

\(\frac{9\sqrt{17}\pi}{4}\) Note: This isn't one of the options given in the question. It appears there's a mistake in the question, as none of these options are correct.

Step-by-step solution

Parameterization of the Circle

The given region R represents a circle with a radius of 3. It can be parameterized using polar coordinates as the following: \(x=r \cos(\theta), y=r \sin(\theta)\), where \(0<=r<=3\) and \(0<=\theta<=2\pi\).

Rewrite function f in Polar Coordinates

Substitute the polar coordinates into the function to simplify the integration process. The function \(f(x, y)\) becomes \(f(r, \theta)=\frac{1}{4} \sqrt{r^{2}} = \frac{r}{4}\)

Apply Surface Area formula

The formula for surface area over the region R in polar coordinates is \(\int\int\sqrt{1+(\frac{{\partial f}}{{\partial r}})^2+(\frac{{r\partial f}}{{\partial \theta}})^2}r\, dr\, d\theta\). Since f(r, theta) does not actually depend on the variable theta, the partial derivatives become \(\frac{{\partial f}}{{\partial r}}=\frac{1}{4}\), and \(\frac{{r\partial f}}{{\partial \theta}}=0\). Substituting these into the surface area formula results in \(\int_0^{2\pi}\int_0^{3}\sqrt{1+\frac{1}{16}}r\, dr\, d\theta\)

Evaluate the Integral

To solve the integral, first simplify it to obtain \(\frac{9\sqrt{17}}{8}\int_0^{2\pi}d\theta\). Evaluate this to get the surface area as \(\frac{9\sqrt{17}\pi}{4}\)