Volume of Solid Revolving Around x & y Axis

[Korupt]

I have to find the volume of the solid whose area is bound by the equations y=-x+3 \ and \ y=x^2-3x as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:

M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}
M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}
A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}
So the center of mass coordinates are:
\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1
\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}
Revolving around x:
d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}
V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083

Revolving around y:
d=(2\pi)(1) = 2\pi
V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206

However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.

 

[SammyS]

I have to find the volume of the solid whose area is bound by the equations y=-x+3 \ and \ y=x^2-3x as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:
M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}
M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}
A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}
So the center of mass coordinates are:
\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1
\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}
Revolving around x:
d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}
V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083
Revolving around y:
d=(2\pi)(1) = 2\pi
V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206
However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.

First of all: Are you instructed to use Pappus's Theorem ?

What precisely does Pappus's Theorem state?

No matter what method you use, this problem includes some tricky details.

 

[Korupt]

I have not been instructed to use any particular method, I just need to find a solution to the problem. For these types of problems I generally use disks or washers to find the volume, but I guess the "tricky" part of this problem is that the bound area crosses the the axis on which it's supposed to rotate, as u can see:
http://www4c.wolframalpha.com/Calculate/MSP/MSP3271a035bi2ii1400fa000012ie2ia8673b6ch5?MSPStoreType=image/gif&s=50&w=381&h=306&cdf=Coordinates&cdf=Tooltips

Pappus's second theorem states that the volume of a solid of revolution generated by the revolution of a region about an external axis is equal to the product of the area of the region and the distance traveled by the region's geometric centroid.

http://mathworld.wolfram.com/PappussCentroidTheorem.html

 

[SammyS]

I have not been instructed to use any particular method, I just need to find a solution to the problem. For these types of problems I generally use disks or washers to find the volume, but I guess the "tricky" part of this problem is that the bound area crosses the the axis on which it's supposed to rotate, as u can see:

Yes, this is the tricky part … but it's easily handled. The trickier part is to revolve this around the y-axis. Only part of the area on the left is overlapped by the area on the right.

Pappus's second theorem states that the volume of a solid of revolution generated by the revolution of a region about an external axis is equal to the product of the area of the region and the distance traveled by the region's geometric centroid.
http://mathworld.wolfram.com/PappussCentroidTheorem.html

The part of Pappus's Theorem you seem to be ignoring is: That the axis must be an external axis to the region being rotated. So, when you calculate the centroid, only include the portion of the region above the x-axis . For the other part only the portion to the right of the y-axis , then do the little portion (that's not overlapped) left of the y-axis.

 

[Korupt]

Yes, this is the tricky part … but it's easily handled. The trickier part is to revolve this around the y-axis. Only part of the area on the left is overlapped by the area on the right.

The part of Pappus's Theorem you seem to be ignoring is: That the axis must be an external axis to the region being rotated. So, when you calculate the centroid, only include the portion of the region above the x-axis . For the other part only the portion to the right of the y-axis , then do the little portion (that's not overlapped) left of the y-axis.

Yeah you are right, it is only external. I figured it out by cutting it into smaller parts like you said and then using disks and washers. Thank you for your help.