anonymous 5 years ago Find the total surface area of the region bounded by the surfaces x2+y2 = 1, x2+y2 = z − 7, and z = 0.

1. anonymous

Swiech's calc 3 class?

2. anonymous

$z = x ^{2}+y ^{2}+7$ $=> \sqrt{1+ z ^{2}_{x} +z ^{2}_{y}} = \sqrt{1+ 4x^{2} + 4y^{2}}$ required surface area = double integral of $\sqrt{1+4x ^{2}+4y ^{2}}$ over the circle $x ^{2}+y ^{2} = 1$ changing to polar co-ordinates $x=rcos \theta, y=rsin \theta , 0\le \theta \le2\pi, 0\le r \le1$ we get the surface area as$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}(\sqrt{1+4r ^{2}}) rdr d \theta$ =$2\pi \int\limits_{0}^{1}r \sqrt{(1+4r ^{2})} dr$ which can be solve very easily. I avoided using vectors

3. anonymous

So Bimol describe what is happenng in the plane. x^2+y^2=1 is a circle but how does that relate to the other one.