surface area

[logistic_guy]

Find the surface area of the given surface. The portion of the cone \(\displaystyle z = \sqrt{x^2 + y^2}\) below the plane \(\displaystyle z = 4\).

 

[khansaheb]

Find the surface area of the given surface. The portion of the cone \(\displaystyle z = \sqrt{x^2 + y^2}\) below the plane \(\displaystyle z = 4\).

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

If you have followed my surface integral problems, you would have already known that we have method 3which I call simple calculus approach.

First I will define the portion to be \(\displaystyle z = f(x,y) = \sqrt{x^2 + y^2}\)

Then

\(\displaystyle \iint\limits_S \ dS = \int\int \sqrt{(f_x)^2 + (f_y)^2 + 1} \ dx \ dy\)

\(\displaystyle = \int_{-4}^{4}\int_{-\sqrt{16 - y^2}}^{\sqrt{16 - y^2}} \sqrt{\frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2} + 1} \ dx \ dy\)

\(\displaystyle = \int_{-4}^{4}\int_{-\sqrt{16 - y^2}}^{\sqrt{16 - y^2}} \sqrt{2} \ dx \ dy\)

I don't want to be fancy this time, so I will move to the polar coordinate

\(\displaystyle = \sqrt{2}\int_{0}^{2\pi}\int_{0}^{4} r \ dr \ d\theta = \sqrt{2}\int_{0}^{2\pi} \frac{r^2}{2}\bigg|_{0}^{4}\ d\theta\)

\(\displaystyle =2\pi\sqrt{2} \ \frac{4^2}{2} = 16\pi\sqrt{2} \approx 71.1\)