Washer Method Cont.

[Silvanoshei]

Revolving about the x-axis for volume.

$\displaystyle y=x^{2}+1, y=(x+3)$

So... R is top x+3 and inner radius the other?

$\displaystyle \pi∫_{1}^{3}(x+3)^{2}-(x^{2}+1)^{2}dx$

Is this the correct setup? I would get something like...

$\displaystyle \pi∫_{1}^{3}(x^{2}+3x+3x+9)-(x{4}+x{2}+x{2}+1)dx?$

 

[DrPhil]

Revolving about the x-axis for volume.

$\displaystyle y=x^{2}+1, y=(x+3)$

So... R is top x+3 and inner radius the other?

$\displaystyle \pi∫_{1}^{3}(x+3)^{2}-(x^{2}+1)^{2}dx$

Is this the correct setup? I would get something like...

$\displaystyle \pi∫_{1}^{3}(x^{2}+3x+3x+9)-(x{4}+x{2}+x{2}+1)dx?$

First thing is the limits of x between where the two curves cross. I seem to find those to be -1 and +2.

I also think the outer radius will be y=(x+3) and the inner will be y=(x^2 + 1), which is the way you wrote it in the formula. So I recommend that you reconsider the limits of integration, and carry on.

 

[Silvanoshei]

Gotcha, I was looking at the Y-Axis for some dumb reason. I agree that -1 and 2 intersect on x-axis.

$\displaystyle v=\pi∫_{-1}^{2}(x^{2}+6x+9)-(x^{4}+2x^{2}+1)dx$

$\displaystyle v=\pi\left[(\frac{x^{3}}{3}+\frac{6x^{2}}{2}+9x)-(\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x)\right]_{-1}^{2}$

$\displaystyle v=(\frac{284\pi}{15})-(\frac{-67\pi}{15})$

$\displaystyle v=\frac{117\pi}{5} ?$

 

[DrPhil]

Gotcha, I was looking at the Y-Axis for some dumb reason. I agree that -1 and 2 intersect on x-axis.

$\displaystyle v=\pi∫_{-1}^{2}(x^{2}+6x+9)-(x^{4}+2x^{2}+1)dx$

$\displaystyle v=\pi\left[(\frac{x^{3}}{3}+\frac{6x^{2}}{2}+9x)-(\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x)\right]_{-1}^{2}$

$\displaystyle v=(\frac{284\pi}{15})-(\frac{-67\pi}{15})$

$\displaystyle v=\frac{117\pi}{5} ?$

On my first attempt at the integral I got the same answer you did! Could be correct.

 

[Silvanoshei]

Nice! Thank you. Hey Dr.Phil, how can I easily tell when I need to use Washer Method or the Shell Method?