Thread: Integrals using cylindrical shells when rotated on non-axis line

1. Integrals using cylindrical shells when rotated on non-axis line

I am getting a bit confused here.. the question asks to find the volume of the solid that results from rotating $y^2=8x, x=2$ around the line $x=4$.

My setup is this: $2 \pi \int_{-4}^4(2-x)(\sqrt{8x}) dx$

but when i integrate and substitute i cant get values for the negative lower limit.. what's my error?

2. Try integratinf from 0-4 and multiply by another constant of 2.

3. i dont understand??

4. we want the volume the green volume as the picture shows us

the function is $y^2=8x$

$y = \sqrt{8x}$

to make it easier let new x=x+4 to make the volume around the y-axis

so the new question will be

$y^2 = 8(x+4)$ and x=-2 around the y-axis

so the picture will be like this

and we want the volume revolution around y-axis
note that $y = \sqrt{8(x+4)}$ is not all the curve if we calculate the volume for y it will be half of what we need since it is symmetric
see this

the volume of this around y-axis

$\int _{-4}^{-2} 2\pi x \sqrt{8(x+4)} dx$

multiply it with 2 to get what you need

any questions

5. or in other way it will equal

$\int_{-4}^{4} \pi \left( \frac{y^2}{8} -4 \right)^2 - 4 dy$

6. Originally Posted by Amer
we want the volume the green volume as the picture shows us

the function is $y^2=8x$

$y = \sqrt{8x}$

to make it easier let new x=x+4 to make the volume around the y-axis

so the new question will be

$y^2 = 8(x+4)$ and x=-2 around the y-axis

so the picture will be like this

and we want the volume revolution around y-axis
note that $y = \sqrt{8(x+4)}$ is not all the curve if we calculate the volume for y it will be half of what we need since it is symmetric
see this

the volume of this around y-axis

$\int _{-4}^{-2} 2\pi x \sqrt{8(x+4)} dx$

multiply it with 2 to get what you need

any questions
Thanks, but how did u get those limits??

7. limits is x=-4 and x=-2 since the region is bounded by this two lines

Cylindrical shell says the volume generated by revolving a region bounded above by a continuous function and from the left by the line x=a and from the right by the line x=b given by

$v= \int_{a}^{b} \pi x f(x) dx$

in the second way I used washers
look to the picture

if we look to the two curves

$y^2 = 8(x+4) \Rightarrow x_1 = \frac{y^2}{8} - 4$

and

$x_2=-2$

want the volume bounded by this to curves revolving around y-axis

find the point of intersection

$-2 = \frac{y^2}{8} - 4 \Rightarrow y=\mp 4$

so the volume will be

$\pi \int_{-4}^{4} (x_1)^2 - (x_2)^2 dy$

$\pi \int_{-4}^{4} \left( \frac{y^2}{8} -4 \right)^2 - (-2)^2 dy$