Find the area of the part of the surface $ z = x^2 - y^2$ which lies in the first octant that is inside cylinder $x^2 + y^2 = 1$

Question

Here is what I have tried so far $S = \int \int \space \sqrt{4z^2 + 4y^2} dA $ would I then change to polar coordinates and what would the bounds for this double integral be ?

Answer 1

To be fair I can't make a sense of $S = \int \int \space \sqrt{4z^2 + 4y^2} dA $, but here's how you can solve the problem. Note that $z$ is explicitly given as a formula of $x,y$, hence you have:

$$A = \iint_S dA = \iint_S\sqrt{1 + z_x^2 + z_y^2}dxdy = \iint_S \sqrt{1 + 4x^2 + 4y^2} dxdy$$

Now note that $x,y$ can take any value inside the disk $x^2 + y^2 = 1$. Sketching the graph can help you to conclude this. Then change to polar coordiantes to get:

$$A = \int_0^1 \int_0^{2\pi} r\sqrt{1+4r^2} d\theta dr = 2\pi \int_1^5 \frac 18\sqrt{x} dx = \frac{\pi}4 \cdot \frac 23 \sqrt{x^3} \bigg|_1^5 =\frac{\pi}6 (5\sqrt{5} - 1)$$

Obviously we used the substitution $x=1 + 4r^2$