integral question calc?
1) How do I set up an integral that gives the area of the surface obtained by rotating the curve y=x^2 on the interval 0<=x<=sqrt2 about the y axis
2) After that, how would I evaluate the integral to find the area?
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Express it as x = sqrt(y)
The integral you seek is the integral of 2(pi)x sqrt(1+(dx/dy)^2)) dy from y = 0 to y = 2
The dx/dy = 1/2sqrt(y), so the square of that is 1/4y.
The integrand is therefore 2(pi) sqrt(y) sqrt(1+1/4y) dy, or 2pi sqrt(y+1/4) dy.
The antiderivative is simply 4pi (y+1/4)^(3/2) /3 from y = 0 to 2
4pi(27/8 - 1/8)/3 = 4pi(26/8)/3 = 13pi/3