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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.
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Swiech's calc 3 class?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z = x ^{2}+y ^{2}+7\] \[=> \sqrt{1+ z ^{2}_{x} +z ^{2}_{y}} = \sqrt{1+ 4x^{2} + 4y^{2}}\] required surface area = double integral of \[\sqrt{1+4x ^{2}+4y ^{2}}\] over the circle \[x ^{2}+y ^{2} = 1\] changing to polar coordinates \[x=rcos \theta, y=rsin \theta , 0\le \theta \le2\pi, 0\le r \le1\] 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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.
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