Question

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

Answer 1

Swiech's calc 3 class?

Answer 2

\[z = x ^{2}+y ^{2}+7\]
\[=> \sqrt{1+ z ^{2}_{x} +z ^{2}_{y}} = \sqrt{1+ 4x^{2} + 4y^{2}}\]
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

Answer 3

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.