- #1

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## Homework Statement

Here is the problem:

## The Attempt at a Solution

I found the intersection point to be (1.10,2.78). Now I don't know what to use for the lower limit.

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- Thread starter niravana21
- Start date

- #1

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Here is the problem:

I found the intersection point to be (1.10,2.78). Now I don't know what to use for the lower limit.

- #2

jhae2.718

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What method of finding volume are you using? Cross sections or cylindrical shell? If using cross sections this will be a dx problem; if using the shell method this will be a dy problem. For this problem I would use the cross section/disk method and thus use x-coordinates as limits of integration. For a, this will be a washer.

- #3

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Edit: so from 0 to 1.10?

- #4

jhae2.718

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It is zero--since we are bounded by the y-axis and the intersection, our limits of integration are 0,1.10.

When we find the volume we are going to add cross sections of the curve from 0 until we reach 1.10. Think of a layer cake--the bottom is zero and the top is the intersection of the curve. Turn it sideways and let the axis down the middle be the x-axis, and that is what we are doing.

When we find the volume we are going to add cross sections of the curve from 0 until we reach 1.10. Think of a layer cake--the bottom is zero and the top is the intersection of the curve. Turn it sideways and let the axis down the middle be the x-axis, and that is what we are doing.

Last edited:

- #5

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And what about the second part?

- #6

jhae2.718

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Use the same limits of integration, but a different cross section. The area of a square is A=s

For a, remember to treat it as a washer. V=pi*S

- #7

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- #8

jhae2.718

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Earlier I accidently used 2.78 instead of 1.10...fixed my above post.

- #9

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thank you very much for all the help you have provided.

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