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

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 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

I'm getting $\displaystyle\frac{45\pi}{2}\approx70.686$

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

For this one, I'm getting $\displaystyle\frac{128\pi}{5}\approx80.425$

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

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?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 $\displaystyle\int_{x=0}^{4}2\,\pi\,x\sqrt{x}\,dx$

I have a better idea ... show howThe 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 did the following:I have a better idea ... show howyouevaluated the integrals so maybe someone can spot your mistake(s).

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

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