# Application with Shell Integration Method

## Homework Statement

Use the shell method to find the volume of the solid obtained by rotating the region bounded by y=x2, y=0 and x=1 about the x-axis.

lim Ʃ2∏RhΔw
Δw->0

## The Attempt at a Solution

I realize this is hard to visualize without a graph. I uploaded how I split the graph.

Solved for y and found y=0, 1.

lim Ʃ2∏RhΔw
Δw->0
=lim Ʃ2∏y(1-y)Δy // made it 1-y because y=x2.
Δy->0
=lim Ʃ2∏(y-y2)Δy
Δy->0
=2∏∫y-y2dy from 0->1
=2∏[y2/2-y3/3] from 0->1
=∏/3

Using Fundamental Theorem of Calculus, volume was found to be ∏/3. However this answer is incorrect. The answer is suppose to be ∏/5. I solved the question initially using the washer method and obtained the answer of ∏/5 but this question asks specifically to use shell method. I don't know how I can fix what I did wrong in this question. I assume I set the wrong h value. I also always get confused on what variable to use x or y, when taking into about if the rotation is about the x or y axis.

Any help would be great thanks!

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LCKurtz
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## Homework Statement

Use the shell method to find the volume of the solid obtained by rotating the region bounded by y=x2, y=0 and x=1 about the x-axis.

lim Ʃ2∏RhΔw
Δw->0

## The Attempt at a Solution

I realize this is hard to visualize without a graph. I uploaded how I split the graph.

Solved for y and found y=0, 1.

lim Ʃ2∏RhΔw
Δw->0
=lim Ʃ2∏y(1-y)Δy // made it 1-y because y=x2.

The quantity in parentheses is xright-xleft. You have (1-y). Now ##x_{right}=1## alright, but since ##y=x^2##, ## x_{left} =\sqrt y## so you should have ##(1-\sqrt y)## there. That will fix it.