1. Surface area question

Question:
x = 1/3 * (y^2 + 2)^(3/2), 1 <= y <= 2

21/2 * pi

My work:
SA = 2pi * integral y ds = 2pi*integral (y * sqrt(1 + (dx/dy)^2)) dy
x = 1/3 * (y^2 + 2)^(3/2)
dx/dy = y*sqrt(y^2 + 2)

Then I'm stuck. Wolfram also says my answer is wrong. I used y variables and the y limits of integration and the answer differs from that of the back of the book. (http://www.wolframalpha.com/input/?i=integrate(sqrt(1+%2B+y^4+%2B+y^2)*y)+*+2pi+from +1+to+2)

Any help would be GREATLY appreciated!

2. Originally Posted by s3a
Question:
x = 1/3 * (y^2 + 2)^(3/2), 1 <= y <= 2

21/2 * pi

My work:
SA = 2pi * integral y ds = 2pi*integral (y * sqrt(1 + (dx/dy)^2)) dy
x = 1/3 * (y^2 + 2)^(3/2)
dx/dy = y*sqrt(y^2 + 2)

Then I'm stuck. Wolfram also says my answer is wrong. I used y variables and the y limits of integration and the answer differs from that of the back of the book. (http://www.wolframalpha.com/input/?i=integrate(sqrt(1+%2B+y^4+%2B+y^2)*y)+*+2pi+from +1+to+2)

Any help would be GREATLY appreciated!
Surface Area $= \int dS$ about the surface S.

What is dS? Well, in a function only in the XY plain it is

$dS= \sqrt {1+ ( \frac{dx}{dy})^2 } dy$

if $x=f(y)$ and $a \le y \le b$

3. But that's exactly what I did: ds = sqrt(1 + y^2 + y^4) dx

Edit: (I forgot to mention that it is rotating about the x-axis)

4. Originally Posted by s3a
But that's exactly what I did: ds = sqrt(1 + y^2 + y^4) dx
Then you integrated wrong but I think your dS is incorrect as well

$x = \frac{ (y^2 + 2)^{ \frac{3}{2} } }{3}$

$\frac{dx}{dy} = y \sqrt {y^2+2}$

$ds= \sqrt { 1 + y^2 (y^2+2) } = \sqrt { 1 + y^4 + 2y^2 }$

I have a 2 where you don't. Perhaps I'm wrong (i didnt go over it lol) and if I am, then you messed up the integration. If not, then try again with that equation

...The rotating about the y-axis means we have

$S=\int 2 \pi x dS$ recompute with the new dS.

5. I said x-axis but that coefficient of 2 was my mistake; thanks!