Area of a solid generated by revolving a region about an axis.

[Vectus]

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

V_total = V₁ + V₂ + V₃ + V₄
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 = 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.

 

[Mark44]

Nice graph!

Find the area 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...

Edit: I completely misread the problem. The region to be revolved is above the x-axis, below the curve y = x^2, and inside the curve y = 2 - x^2.
[STRIKE]Yes, it's extraneous. The region to be revolved is completely defined by the two parabolas.[/STRIKE]

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?

No. If you have a solid object that takes up space, its volume is a positive number.

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.

I wouldn't do that. When you set up an integral for a volume of revolution, there are two basic approaches. One approach uses disks or washers, disks with a hole in them, and the other uses cylindrical shells, which are a little like the layers of an onion. The easy approach on this problem is to use washers.

Divide the interval [0, 1] along the y-axis into subintervals of length \Delta y. If you take a horizontal strip that runs from the inner parabola (y = x^2) to the outer parabola (y = 2 - x^2), and then rotate it around the y-axis, you get a "washer" whose volume is its cross-section area times \Delta y.

The volume of this typical volume element is pi * [ (outer radius)^2 - (inner radius)^2] * \Delta y. This expression is your integrand, and the limits of integration are 0 and 1.

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/2


V₂ = π∫₀¹ -√(y)² dy
V₂ = -π∫₀¹ y dy
V₂ = -π[ y²/2]₀¹
V₂ = -π/2


V₃ = 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/2


V₄ = 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

V_total = V₁ + V₂ + V₃ + V₄
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 = 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.

 

[Sethric]

A few things:

Make sure you state volume next time, I read through the problem and thought you meant surface area for the first half of the post.

There is a region bounded by all three curves: one with a lower bound of x = 0 and an upper bound of either x^2 or 2-x^2 depending on x coordinate. Try that one; you will need to break it into two parts if you use symmetry.

Symmetry will never give you a zero volume, it will just allow you to compute easier by only integrating part of the problem and then using symmetry as a multiplier.

I noticed your algebra isn't quite correct on some of your integrals. The volume formula has outer radius squared, a larger positive number, minus inner radius squared, a smaller positive number. It should always end up positive.

 

[SteamKing]

The problem said find the area, not the volume, of the solid of revolution. Well, which is it?

 

[Vectus]

A few things:

Make sure you state volume next time, I read through the problem and thought you meant surface area for the first half of the post.

The problem said find the area, not the volume, of the solid of revolution. Well, which is it?

Sorry about that. I could have sworn I typed 'volume' but for some reason it came out 'area.' Fixed now.

 

[Mark44]

Ahh, yes. That makes sense! I didn't see that, because every problem I've worked involving that method had a region with a gap in the center, so the solid was an actual 'shell.' I never visualized it quite like that. Thanks, that really clears this up for me.

Check what I wrote again, since I realized that I was not seeing the correct region. I have edited what I wrote to reflect a better understanding.