Surface area of functions without definite integrals

[Interesting]

Homework Statement

The curve is rotated about the y axis, find the area of the resulting surface.

y=(1/4)X²-.5ln|x| 1<_X<_2

Homework Equations

S=2(pi)(f(x))[tex]\sqrt{}1+f'(x)^2[/tex]

The Attempt at a Solution

Alright I'm not entirely sure where to even begin. Since I'm rotating about the Y-axis I know that I need to solve the limits wrt y, which would be 1/4<_y<_0.65345
Then I need to solve the equation so X is isolated, then take the derivative of that.
I'm not entirely sure where to even begin, I know that I could take everything to the e^ to get rid of the natural log of x, but then everything is to the e^.

 

[Dick]

You can't explicitly write a formula for x(y). But you don't have to. You have to integrate 2*pi*x(y)*sqrt(1+x'(y)^2)*dy. dy=(x/2-1/(2x))*dx. Use that to find dx/dy and convert the integration over dy to an integration over dx. You can manipulate the expression inside the square root into a perfect square.

 

[Interesting]

Thanks for the help, I don't completely follow you though.

You can't explicitly write a formula for x(y). But you don't have to. You have to integrate 2*pi*x(y)*sqrt(1+x'(y)^2)*dy. dy=(x/2-1/(2x))*dx. Use that to find dx/dy and convert the integration over dy to an integration over dx. You can manipulate the expression inside the square root into a perfect square.


1/(x/2-1/(2x))=dx/dy, now that's x'(y), and then take the integral of this function to get X? If I take the integral of that I get 2ln|x|+x^2 = y, but I don't really think this is right because it's almost where I started from, and there aren't any y's in the equation and x isn't isolated. x(y) should equal the integral of dx/dy though? Thanks a lot.

 

[Dick]

No, don't integrate dx/dy. You want to integrate x*sqrt(1+(dx/dy)^2)*dy. That's x*sqrt(1+(dx/dy)^2)*(x/2-1/(2x))*dx. You can write it completely as an integral over x without ever knowing an explicit form for x as a function of y. You are basically just changing the variable of integration from y to x.

 

[Interesting]

Alright I'm still stuck.

so dx/dy= 1/((x/2-1/(2x)) which is the same as 2/x - 2x, so (2/x-2x)(2/x-2x)= 4/x2-8+4x2, take a 4 out and you get 1/x2-2x2+1, get a common denominator and you get (x4-2x2+1)/(x2), factors to (x2-1)(x2-1)/(x2) + 1. I know this is going to be really obvious, but I can not see the perfect square factorization of this.

Thanks a lot for the help, the book we have doesn't offer very good explanations.

 

[Dick]

1/(x/2-1/(2x)) is NOT the same as 2/x-2x. Now you are just being sloppy with algebra. Try again.

 

[Interesting]

Okay, just to check in, the solution to the square root should be (x2+1)/(x2-1) (providing my algebra is correct, which it most likely isn't). Then I multiply that solution by (x2/2 - 1/2) (by multiplying the x into dy). Then I get two integrals, and then I long divide and get a really long integral and another long one? (Just checking so I don't waste an hour going down the wrong track).

 

[Dick]

I think your algebra is correct this time. (x^2/2-1/2)=(x^2-1)/2. Looks to me like you've got some cancellation there when you multiply by (x^2+1)/(x^2-1). May not be nearly as hard as you are thinking.

 

[Interesting]

Ohhh good call! I got a final answer of 12pi, which is simple enough to make me think it's the correct one! Thanks Dick!