surface area of the part of the circular paraboloid

Question

Find the surface area of the part of the circular paraboloid that lies inside the cylinder

Answer 1

You do this exactly the same way as your other surface area problem. Recall the surface area formula:

$$f_y=2y, f_x=2x.$$

So we have $$S=\int_{-5}^5\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}\sqrt{(2x)^2+(2y)^2+1}dydx$$

Can you take it from there?

Alternatively, you could do this using polar coordinates, which is much easier:

$$S=\int_{0}^{2\pi}\int_{0}^{5}\sqrt{4r^2+1}rdrd\theta$$