Find the area of the surface generated when the given curve is revolved about the y-axis.

The part of the curve y=4x-4 between the points ((3/2),2) and ((7/2),10)Leave the answer in pi

I can't afford money to pay you! But I would appreciate help!

K so here is what you do, first notice you are revolving around the y axis, so you need it to be in terms of y.y=4x-4x=(y+4)/4now that you have the function you need to recall the formulaSA = integral from A to B of 2pi * f(y) * dsds = arc length known as sqrt(1+f'(Y)^2) dySo now lets find f'(y), simply its the derivative, which in this case is 1/4plug in all your formulas you get something likethe integral from 2 to 10 (since we are in terms of y) of 2pi * (y+4)/4 * sqrt(1+(1/4)^2) dypull the 2pi out in front, and the simplify whats inside the sqrt and distribute, we get2pi the integral from 2 to 10 of sqrt(17/16)y/4 + sqrt(17/16) dynext you integrate, using power rule.2pi the integral from 2 to 10 of sqrt(17/16)y^2/8 + sqrt(17/16)ylastly you plug everything in,2pi((sqrt(17/16)10^2/8 + sqrt(17/16)10) - (sqrt(17/16)2^2/8 + 2sqrt(17/16))simplifyyou get your final answer to be10sqrt(17)*pi

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