- #1
Vectus
- 46
- 0
Find the volume of the solid generated by revolving the region bounded by y = x2, y = 2-x2, 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
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.
V1 = π∫01 √(y)2 dy
V1 = π∫01 y dy
V1 = π[ y2/2]01
V1 = π/2V2 = π∫01 -√(y)2 dy
V2 = -π∫01 y dy
V2 = -π[ y2/2]01
V2 = -π/2V3 = π∫12 √(2 - y)2 dy
V3 = π∫12 2 - y dy
V3 = π[2y - y2/2]12
V3 = π[ ( 2(2) - (2)2/2 ) - ( 2(1) - (1)2/2 ) ]
V3 = π[ ( 2 ) - ( 3/2 ) ]
V3 = π/2V4 = π∫12 -√(2 - y)2 dy
V4 = -π∫12 2 - y dy
V4 = -π[2y - y2/2]12
V4 = -π[ ( 2(2) - (2)2/2 ) - ( 2(1) - (1)2/2 ) ]
V4 = -π[ ( 2 ) - ( 3/2 ) ]
V4 = -π/2
Vtotal = V1 + V2 + V3 + V4
Vtotal = π/2 - π/2 + π/2 - π/2
Vtotal = 0
Which is what my intuition told me by looking at the graph. Is this correct? Or should it be: Vtotal = V1 + V3 = 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.
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
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.
V1 = π∫01 √(y)2 dy
V1 = π∫01 y dy
V1 = π[ y2/2]01
V1 = π/2V2 = π∫01 -√(y)2 dy
V2 = -π∫01 y dy
V2 = -π[ y2/2]01
V2 = -π/2V3 = π∫12 √(2 - y)2 dy
V3 = π∫12 2 - y dy
V3 = π[2y - y2/2]12
V3 = π[ ( 2(2) - (2)2/2 ) - ( 2(1) - (1)2/2 ) ]
V3 = π[ ( 2 ) - ( 3/2 ) ]
V3 = π/2V4 = π∫12 -√(2 - y)2 dy
V4 = -π∫12 2 - y dy
V4 = -π[2y - y2/2]12
V4 = -π[ ( 2(2) - (2)2/2 ) - ( 2(1) - (1)2/2 ) ]
V4 = -π[ ( 2 ) - ( 3/2 ) ]
V4 = -π/2
Vtotal = V1 + V2 + V3 + V4
Vtotal = π/2 - π/2 + π/2 - π/2
Vtotal = 0
Which is what my intuition told me by looking at the graph. Is this correct? Or should it be: Vtotal = V1 + V3 = 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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