Integration By Volume

1. Feb 23, 2013

kmr159

1. Find the volume for the solid obtained by rotating the region bounded by the given curves about the specified line. y=(1/4)(x^2), x = 2, y = 0, about the y axis

2. Use the disk/washer method of integration to find the volume

3. When I try to solve this problem using small changes in x (dx) I get the wrong answer. Method: Take integral from 0 to 2 of a function for the area of a circle.

∫ from 0 to 2 : 2[(∏)*((x^2)/8)^2] dx → ∫ from 0 to 2 : (∏x^4)/32 → ∏/5

I do not see why I need to use the washer method as using the disk method encounters no hollow space

Please tell what I'm doing wrong.

Thanks

I'm not comfortable with the washer method, that's why I solved the problem with small change in the x axis.

2. Feb 23, 2013

Dick

You appear to be rotating around the x-axis. The problem asks you to rotate around the y-axis. Then there is a hollow space between x=0 and x=sqrt(4y). The part between x=sqrt(4y) and x=2 is the filled in part.

3. Feb 23, 2013

eumyang

If you're supposed to rotate around the y-axis and use the disk/washer method, then you need to integrate with respect to y, not x. Rewrite $y = \frac{1}{4}x^2$ in terms of x. Use x = 2 to find the upper limit of integration in terms of y. Also, this integral should be multiplied by π, not 2π.

EDIT: Beaten to it. ;)

4. Feb 23, 2013

Dick

Oh, not by much. Both posts have valuable information.

5. Feb 23, 2013

kmr159

I am rotating about the y axis

I am multiplying the entire ∏r2 portion by 2 because my equation only finds the area one half of the resultant solid

to find r I divided (1/4x2) by 2 to find the radius of my circles.

thanks again

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6. Feb 24, 2013

eumyang

No, you don't need to do that! The problem, as you stated it, says to
The resulting area of the region lies only in the 1st quadrant, and that is the only region that is revolving around the y-axis. There is no need to "multiply by 2."

I don't understand what you are saying here. You need to use the washer method, so you have to find two radii, normally notated as R and r. The integrand would contain R2 - r2, and pi is outside of the integral.