The portion of the paraboloid z=9-x^2-y^2 between the cylinders x^2+y^2=1 and x^2+y^2=4
and please show working out.
and please show working out.
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I'll give you a hint. See that EVERYTHING here is symmetric around the origin, so a good start is to convert to polar coordinates.
In polar coordinates, the problem becomes to find the surface area of the paraboloid
z = 9 - r^2
subject to the constraint that
1 <= r <= 2
Be careful, you will still need to integrate here. What is the area? You must integrate an area element that looks like
sqrt(1 + (dz/dr)^2) * 2*pi*r dr
Since dz/dr = -2r, we need to do the integration
int(sqrt(1 + 4*r^2) *2*pi*r dr) = (4*r^2 + 1)^(3/2) *2*pi/3
Evaluated between limits of 1 and 2, the definite integral is
(17^(3/2) - 5^(3/2)) * 2*pi/3 = 123.38595878857
In polar coordinates, the problem becomes to find the surface area of the paraboloid
z = 9 - r^2
subject to the constraint that
1 <= r <= 2
Be careful, you will still need to integrate here. What is the area? You must integrate an area element that looks like
sqrt(1 + (dz/dr)^2) * 2*pi*r dr
Since dz/dr = -2r, we need to do the integration
int(sqrt(1 + 4*r^2) *2*pi*r dr) = (4*r^2 + 1)^(3/2) *2*pi/3
Evaluated between limits of 1 and 2, the definite integral is
(17^(3/2) - 5^(3/2)) * 2*pi/3 = 123.38595878857