Question 1011000: Find the surface area of the cone z=sqrt(x^2+y^2) below the plane z=8.
Please show your solution step by step.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We want the surface area of the portion of the cone z^2 = x^2 + y^2 between z=0 and z=8. The equation of the cone in cylindrical coordinates is just z = r, so we can take as our parameters r and t (representing theta).
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treat that potion(S) of the cone as a graph whose shadow D on the xy-plane is the disk of radius 8. To convert from dS to dA, we need to calculate
sqrt(1 + (partial derivative z / partial derivative x)^2 + (partial derivative z / partial derivative y)^2) = sqrt(1 + (x^2/(x^2+y^2)) + (y^2/(x^2+y^2))) = sqrt(2)
therefore the surface area is a double integral
integrate t from 0 to 2pi integrate r from 0 to 8 sqrt(2)*dr*dt=
64*pi*sqrt(2)
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alternately,
We can unwrap the cone to a sector of a circular disk, with radius 8*sqrt(2) and outer circumference 16*pi (compared to 16*pi*sqrt(2) for the whole circle), so the surface area is pi(8*sqrt(2))^2 / sqrt(2) = 64*pi*sqrt(2)
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