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**Find the volume of the solid generated by revolving the region bounded by y = x**

^{2}, y = 2-x^{2}, and x = 0 about the y-axis.I'm taking Calc I and this question was on a test I took yesterday. A couple things about it didn't make much sense, which suggests to me that I'm missing something. I actually got hung up on it and ran out of time before I finished solving it.

**Attempted solution:**

The first thing I did was graph the equation and find points of intersection.

[PLAIN]http://img27.imageshack.us/img27/8804/graphlp.jpg [Broken]

First off, the boundary x=0 appears to be extraneous. This doesn't make sense, because my instructor usually does not include any extraneous information on her tests. But ignoring this for now...

From this, I wasn't really sure how to proceed. The problem says to revolve this area around the y-axis.. and since the region is symmetrical with respect to the y-axis, wouldn't the volume of this solid be 0? This really doesn't make sense. Again, my instructor is not one to put trick problems on tests.

I could find the area of the region trivially using the fundamental theorem of calculus. I actually did this on the test without really thinking it through, assuming it would be a problem where I'd have to integrate the area over an interval to get the volume, or use a geometric formula to find the volume of the solid. But since the solid generated isn't a geometric shape I'm familiar with, that doesn't help. This is where I got stuck.

But if I'm not mistaken, I should be able to split up the region into four parts. This would simplify it into something I know how to work.

The four regions would be bounded by the line x=1, the y-axis, and:

x = √(y), x = -√(y), x = √(2-y), and x = -√(2 - y).

Then the volume of the solid would be equivalent to the sum of the volumes of the solids generated by rotating each of the new regions about the y-axis.

V

_{1}= π∫

_{0}

^{1}√(y)

^{2}dy

V

_{1}= π∫

_{0}

^{1}y dy

V

_{1}= π[ y

^{2}/2]

_{0}

^{1}

V

_{1}= π/2

V

_{2}= π∫

_{0}

^{1}-√(y)

^{2}dy

V

_{2}= -π∫

_{0}

^{1}y dy

V

_{2}= -π[ y

^{2}/2]

_{0}

^{1}

V

_{2}= -π/2

V

_{3}= π∫

_{1}

^{2}√(2 - y)

^{2}dy

V

_{3}= π∫

_{1}

^{2}2 - y dy

V

_{3}= π[2y - y

^{2}/2]

_{1}

^{2}

V

_{3}= π[ ( 2(2) - (2)

^{2}/2 ) - ( 2(1) - (1)

^{2}/2 ) ]

V

_{3}= π[ ( 2 ) - ( 3/2 ) ]

V

_{3}= π/2

V

_{4}= π∫

_{1}

^{2}-√(2 - y)

^{2}dy

V

_{4}= -π∫

_{1}

^{2}2 - y dy

V

_{4}= -π[2y - y

^{2}/2]

_{1}

^{2}

V

_{4}= -π[ ( 2(2) - (2)

^{2}/2 ) - ( 2(1) - (1)

^{2}/2 ) ]

V

_{4}= -π[ ( 2 ) - ( 3/2 ) ]

V

_{4}= -π/2

V

_{total}= V

_{1}+ V

_{2}+ V

_{3}+ V

_{4}

V

_{total}= π/2 - π/2 + π/2 - π/2

V

_{total}= 0

Which is what my intuition told me by looking at the graph. Is this correct? Or should it be: V

_{total}= V

_{1}+ V

_{3}= π? The whole negative volume thing really messes with my head on some levels, but it seems to me like the first answer is correct. However, as I stated, it seems uncharacteristic of my instructor to give a problem like that, and it's possible I'm wrong and the problem should be interpreted to exclude negative volume, sort of like some 'area between functions' problems.

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