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Calculus/rotating about an axis and finding the area

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Expert: - 9/14/2008


Question
The given curve is rotated about the y-axis. Find the area of the resulting surface.

y = 1/4 x^2 - 1/2 ln(x)
0 ≤ x ≤ 2


and how it is different in solving:

Find the area of the surface obtained by rotating the curve about the x-axis.
y = sqrt(2 + 2 x)
3≤x≤11


thank you so much if you can help, i am just completely lost.

Answer
They are different because they are entirely different functions.

To rotate a function around the y axis means you have to integrate 2πxf(x).

In the first case it is integral(2π(x^3 /4 - xln(x)/2)dx from 0 to 4.
The 2π can be factored out, the integral of x^3 /4 is 4x^4 / 16.  The integral of xln(x)/2 can be found by u-v.

Let u(x)=ln(x), so du=x/2  and  du=dx/x, so v=x².  
Evaluate uv - integral(v du).
You get x²ln(x) - integral(x^3/2)(from 0 to 4).  I know you can do a power of x integral, so that's my answer.

The second problem is the integral(sqrt(2+2x))dx from 3 to 11.
Let u=2+2x, the du=2 dx.  This is now the integral(u) from (8 to 24).
I get the 8 and 24 since u=2+2x, x is 3 and 11.

That's now a straightforward integral as well - the answer is
u²/2 (from 8 to 24).

Note when I say from(a to b), a is on the bottom of the integral and b is on the top.

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