Surface Area of a Solid of Revolution

[Ki-nana18]

Homework Statement

Find the area of the surface generated when you rotate the parabola y=x² 0 less than or equal to x less than or equal to the square root of k, around the y-axis. You should end up with a simple formula in terms of the constant k.

Homework Equations

S=2[tex]\pi[/tex][tex]\int[/tex]yds

The Attempt at a Solution

I suspect that the simple formula is the volume of a sphere. I got all the way to applying the fundamental theorem of calculus and so far I have 2pi[((12(square root of k)+3)/(18))^(3/2)-(1/12)]

 

[LCKurtz]

Homework Statement

Find the area of the surface generated when you rotate the parabola y=x² 0 less than or equal to x less than or equal to the square root of k, around the y-axis. You should end up with a simple formula in terms of the constant k.

Homework Equations

S=2[tex]\pi[/tex][tex]\int[/tex]yds

The Attempt at a Solution

I suspect that the simple formula is the volume of a sphere. I got all the way to applying the fundamental theorem of calculus and so far I have 2pi[((12(square root of k)+3)/(18))^(3/2)-(1/12)]

Your setup is incorrect because the radius of rotation should be x, not y. But I'm curious why you would think the answer for surface area would give a volume? And of a sphere?

 

[Ki-nana18]

Well I figured since I am rotating around the y-axis I would have the equation set up as x=[tex]\sqrt{}y[/tex].

 

[LCKurtz]

Your setup is incorrect because the radius of rotation should be x, not y. But I'm curious why you would think the answer for surface area would give a volume? And of a sphere?

Well I figured since I am rotating around the y-axis I would have the equation set up as x=[tex]\sqrt{}y[/tex].

When you rotate about the y-axis the radius is in the x direction, your equation is given in terms of x, and the natural variable to use is x.