Mathematics Tutor: None Selected Time limit: 1 Day

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)

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

Feb 25th, 2015

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-4
x=(y+4)/4

now that you have the function you need to recall the formula
SA = integral from A to B of 2pi * f(y) * ds
ds = arc length known as sqrt(1+f'(Y)^2) dy

So now lets find f'(y), simply its the derivative, which in this case is 1/4

plug in all your formulas you get something like

the integral from 2 to 10 (since we are in terms of y) of 2pi * (y+4)/4 * sqrt(1+(1/4)^2) dy

pull the 2pi out in front, and the simplify whats inside the sqrt and distribute, we get

2pi the integral from 2 to 10 of sqrt(17/16)y/4 + sqrt(17/16) dy

next you integrate, using power rule.

2pi the integral from 2 to 10 of sqrt(17/16)y^2/8 + sqrt(17/16)y

lastly you plug everything in,

2pi((sqrt(17/16)10^2/8 + sqrt(17/16)10) - (sqrt(17/16)2^2/8 + 2sqrt(17/16))

simplify

10sqrt(17)*pi

Feb 25th, 2015

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Feb 25th, 2015
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Feb 25th, 2015
Dec 11th, 2016
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