Answer:
About the x axis
[tex] V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}[/tex]
About the y axis
[tex] V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}[/tex]
About the line y=8
[tex] V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}[/tex]
About the line x=2
[tex] V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi[/tex]
Step-by-step explanation:
For this case we have the following functions:
[tex] y = 2x^2 , y=0, X=2[/tex]
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:
[tex] A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4[/tex]
And we can find the volume with this formula:
[tex] V = \int_{a}^b A(x) dx[/tex]
[tex] V= 4\pi \int_{0}^2 x^4 dx[/tex]
[tex] V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}[/tex]
About the y axis
For this case we need to find the function in terms of x like this:
[tex] x^2 = \frac{y}{2}[/tex]
[tex] x = \pm \sqrt{\frac{y}{2}}[/tex] but on this case we are just interested on the + part [tex] x=\sqrt{\frac{y}{2}}[/tex] as we can see on the second figure attached.
We can find the area like this:
[tex] A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})[/tex]
And we can find the volume with this formula:
[tex] V = \int_{a}^b A(y) dy[/tex]
[tex] V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy[/tex]
[tex] V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}[/tex]
About the line y=8
The figure 3 attached show the radius. We can find the area like this:
[tex] A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)[/tex]
And we can find the volume with this formula:
[tex] V = \int_{a}^b A(x) dx[/tex]
[tex] V= \pi \int_{0}^2 64-32x^2 +4x^4 dx[/tex]
[tex] V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}[/tex]
About the line x=2
The figure 4 attached show the radius. We can find the area like this:
[tex] A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}[/tex]
And we can find the volume with this formula:
[tex] V = \int_{a}^b A(y) dy[/tex]
[tex] V= \frac{\pi}{2} \int_{0}^8 y dy[/tex]
[tex] V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi[/tex]
