Question

How do you find the volume of the solid generated by revolving the region bounded by the graphs

y=2x^2,y=0,x=2, about the x-axis, y-axis, the line y=8, the line x=2?

Answer 1

About the x axis

`V = 4\pi[ (x^5)/(5)] \Big|_0^2 =4\pi *(32)/(5)= (128 \pi)/(5)`

About the y axis

`V = \pi [4y -y^2 +(y^3)/(12)] \Big|_0^8 =\pi *(32)/(3)= (32 \pi)/(3)`

About the line y=8

`V = \pi [64x -(32)/(3)x^3 +(4)/(5)x^5] \Big|_0^2 =\pi *(128-(256)/(3) +(128)/(5))= (1024 \pi)/(5)`

About the line x=2

`V = (\pi)/(2) [(y^2)/(2)] \Big|_0^8 =(\pi)/(4) *(64)= 16\pi`

Explanation:

For this case we have the following functions:

`y = 2x^2 , y=0, X=2`

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

`A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4`

And we can find the volume with this formula:

`V = \int_(a)^b A(x) dx`

`V= 4\pi \int_(0)^2 x^4 dx`

`V = 4\pi [(x^5)/(5)] \Big|_0^2 =4\pi *(32)/(5)= (128 \pi)/(5)`

About the y axis

For this case we need to find the function in terms of x like this:

`x^2 = (y)/(2)`

`x = \pm \sqrt{(y)/(2)}` but on this case we are just interested on the + part
`x=\sqrt{(y)/(2)}` as we can see on the second figure attached.

We can find the area like this:

`A = \pi r^2 = \pi (2-\sqrt{(y)/(2)})^2 = \pi (4 -2y +(y^2)/(4))`

And we can find the volume with this formula:

`V = \int_(a)^b A(y) dy`

`V= \pi \int_(0)^8 2-2y +(y^2)/(4) dy`

`V = \pi [4y -y^2 +(y^3)/(12)] \Big|_0^8 =\pi *(32)/(3)= (32 \pi)/(3)`

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

`A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)`

And we can find the volume with this formula:

`V = \int_(a)^b A(x) dx`

`V= \pi \int_(0)^2 64-32x^2 +4x^4 dx`

`V = \pi [64x -(32)/(3)x^3 +(4)/(5)x^5] \Big|_0^2 =\pi *(128-(256)/(3) +(128)/(5))= (1024 \pi)/(5)`

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

`A = \pi r^2 = \pi (\sqrt{(y)/(2)})^2 = \pi(y)/(2)`

And we can find the volume with this formula:

`V = \int_(a)^b A(y) dy`

`V= (\pi)/(2) \int_(0)^8 y dy`

`V = (\pi)/(2) [(y^2)/(2)] \Big|_0^8 =(\pi)/(4) *(64)= 16\pi`