surface area of the part of the circular paraboloid Find the surface area of the part of the circular paraboloid!enter image description here that lies inside the cylinder !enter image description here

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

!integral

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

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:

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