Need answer to shells method question checked

Jul 2015
51
0
California


It is somewhat tedious to show the work I did in order to arrive at the answer, so instead I would like somebody to simply tell me if the answer I got to this question is correct:
My answer: V = 268.6061719 cm^3
 
Jul 2015
51
0
California


I attempted to do another shells method problem just now and would like the answer to this one checked as well.
My answer: V = 268.0825731 cm^3
 
May 2015
74
34
USA


It is somewhat tedious to show the work I did in order to arrive at the answer, so instead I would like somebody to simply tell me if the answer I got to this question is correct:
My answer: V = 268.6061719 cm^3
I'm getting $\displaystyle\frac{45\pi}{2}\approx70.686$
 
May 2015
74
34
USA


I attempted to do another shells method problem just now and would like the answer to this one checked as well.
My answer: V = 268.0825731 cm^3
For this one, I'm getting $\displaystyle\frac{128\pi}{5}\approx80.425$
 
Jul 2015
51
0
California
Can I see how you got the answer to either of the problems? Usually if my answer doesn't match up with somebody else's, I'm the one who messed up.
 
May 2015
74
34
USA
Can I see how you got the answer to either of the problems? Usually if my answer doesn't match up with somebody else's, I'm the one who messed up.
1) Volume of shell element:
length is the circumference: $2\,\pi\,r=2\,\pi\,(x+1)$
height: $3x-x^2$
thickness: $dx$

Integral is $\displaystyle\int_{x=0}^{3}2\,\pi\,(x+1)(3x-x^2)dx$

2) Volume of shell element:
length is the circumference: $2\,\pi\,r=2\,\pi\,x$
height: $\sqrt{x}$
thickness: $dx$

Integral is $\displaystyle\int_{x=0}^{4}2\,\pi\,x\sqrt{x}\,dx$
 
Jul 2015
51
0
California
1) Volume of shell element:
length is the circumference: $2\,\pi\,r=2\,\pi\,(x+1)$
height: $3x-x^2$
thickness: $dx$

Integral is $\displaystyle\int_{x=0}^{3}2\,\pi\,(x+1)(3x-x^2)dx$

2) Volume of shell element:
length is the circumference: $2\,\pi\,r=2\,\pi\,x$
height: $\sqrt{x}$
thickness: $dx$

Integral is $\displaystyle\int_{x=0}^{4}2\,\pi\,x\sqrt{x}\,dx$
The integrals you ended up with were the same as mine, so I must have messed up evaluating them somehow. Can I see how you did the evaluation?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
The integrals you ended up with were the same as mine, so I must have messed up evaluating them somehow. Can I see how you did the evaluation?
I have a better idea ... show how you evaluated the integrals so maybe someone can spot your mistake(s).
 
Jul 2015
51
0
California
I have a better idea ... show how you evaluated the integrals so maybe someone can spot your mistake(s).
I did the following:

1. The formula I started out with was: V = 2╥ * ∫ [(x + 1)(3x - x^2)] (when evaluated from 0 to 3)
2. I then multiplied (x + 1) into (3x - x^2) to get: 3 - x^3 + 3x
3. Integrated 3 - x^3 + 3x to get: 3x + (-x^4 / 4) + (3 * (x^2 / 2))
4. The formula from step one is now V = 2╥ * ∫ [3x + (-x^4 / 4) + (3 * (x^2 / 2))] (when evaluated from 0 to 3)
5. Result of the formula for when x = 3 is: 2╥ * 42.75 = 268.6061719. Result of the formula for x = 0 is: 2╥ * 0 = 0.
6. Subtract 0 from 268.6061719 to get: V = 268.6061719
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
algebra mistake to start ...

$(x+1)(3x-x^2) = -x^3+2x^2+3x$

... try again.