Finding Volume by shell method not around the x or y axis

[krice]

Find the volume of the region bounded by the given curves about the specified axis.

y=4x-x² , y= 3 ; about x = 1

This is what i came up with for the Volume formula:

2π₁∫³[(x-1)((4x-x²-3)]dx

I don't understand why i am getting the answer wrong. But I think it might be in the way I'm setting up the formula for the volume of the region.

any insight is appreciated.

Thanks!

 

[DrPhil]

Find the volume of the region bounded by the given curves about the specified axis.

y=4x-x² , y= 3 ; about x = 1

This is what i came up with for the Volume formula:

2π₁∫³[(x-1)((4x-x²-3)]dx

I don't understand why i am getting the answer wrong. But I think it might be in the way I'm setting up the formula for the volume of the region.

any insight is appreciated.

Thanks!

ok. The radius in the formula will be x-1, hence the shell circumference
\(\displaystyle C(x) = 2\pi (x - 1) \).

The other factor to consider is \(\displaystyle \Delta y(x)\). Find the points where the two curves cross, by setting
\(\displaystyle 4x - x^2 = 3 \implies x^2 - 4x + 3 = (x-3)(x-1) = 0\)
You did that right, too, so the integral is

\(\displaystyle \displaystyle V = 2\pi \int_1^3 (x-1)(-x^2 + 4x - 3)\ dx \)

....\(\displaystyle \displaystyle= 2\pi \int_1^3 (-x^3 + 5x^2 - 7x + 3)\ dx \)

That is one step past what you showed. Looks to me like you set up the integral correctly.

 

[krice]

ok. The radius in the formula will be x-1, hence the shell circumference
\(\displaystyle C(x) = 2\pi (x - 1) \).

The other factor to consider is \(\displaystyle \Delta y(x)\). Find the points where the two curves cross, by setting
\(\displaystyle 4x - x^2 = 3 \implies x^2 - 4x + 3 = (x-3)(x-1) = 0\)
You did that right, too, so the integral is

\(\displaystyle \displaystyle V = 2\pi \int_1^3 (x-1)(-x^2 + 4x - 3)\ dx \)

....\(\displaystyle \displaystyle= 2\pi \int_1^3 (-x^3 + 5x^2 - 7x + 3)\ dx \)

That is one step past what you showed. Looks to me like you set up the integral correctly.

Then i must be just making an integration error or possibly just not simplifying properly. Thank you for the answer, the fact that I'm finally understanding how to set up the formula is honestly great news.

 

[krice]

The problem was when i foiled out the formula

For whatever reason I was putting a 4 at the end:

Your correct version:
..=2π∫31(−x3+5x2−7x+3) dx

My incorrect version:
..=2π∫31(−x3+5x2−7x+4) dx