Rotating a Curve & Line Around the X Axis: A Math Problem

[hvidales]

Homework Statement

The curve x=y^(2) and the line x=4 is rotated about the x axis.

Homework Equations

pi* integral from a to b of Radius^(2)

The Attempt at a Solution

pi* integral from 0 to 4 of (square root of x)^(2) dx.

My teacher has this answer as 8pi but I think that that is wrong. Shouldn't it be 16 pi because you have to account for symmetry?

 

[LCKurtz]

Homework Statement

The curve x=y^(2) and the line x=4 is rotated about the x axis.

Homework Equations

pi* integral from a to b of Radius^(2)

The Attempt at a Solution

pi* integral from 0 to 4 of (square root of x)^(2) dx.

My teacher has this answer as 8pi but I think that that is wrong. Shouldn't it be 16 pi because you have to account for symmetry?

Your formula for the volume of the little disk is for the whole disk: $\pi r^2 dx$. That's the area of the circle times the thickness. Your teacher is correct.

 

[hvidales]

I see! How would you do this if you are doing the shell method? I am stuck at this part:

2pi* integral from -2 to 2 of y[4-y^(2)] dy. I am stuck there lol.

 

[LCKurtz]

I see! How would you do this if you are doing the shell method? I am stuck at this part:

2pi* integral from -2 to 2 of y[4-y^(2)] dy. I am stuck there lol.

You almost have it correct. But remember that the $2\pi y$ accounts for revolving the area. You only revolve the area above the $x$ axis. Your original statement of the problem is not well worded in that regard. So $y$ would only vary from 0 to 2. Otherwise the volume is given twice.

 

[hvidales]

I see and thanks. However, how come for this problem: y=x^(2), y=2-x^(2), about x=1 the limits are from -1 to 1.?

 

[LCKurtz]

Remember, the limits are always for the area that is being revolved. Think about what limits you would use if you were just calculating the area.

 

[hvidales]

Thank you!