Answer:
About the x axis
![V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%204%5Cpi%5B%20%5Cfrac%7Bx%5E5%7D%7B5%7D%5D%20%5CBig%7C_0%5E2%20%3D4%5Cpi%20%2A%5Cfrac%7B32%7D%7B5%7D%3D%20%5Cfrac%7B128%20%5Cpi%7D%7B5%7D)
About the y axis
![V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B4y%20-y%5E2%20%2B%5Cfrac%7By%5E3%7D%7B12%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cpi%20%2A%5Cfrac%7B32%7D%7B3%7D%3D%20%5Cfrac%7B32%20%5Cpi%7D%7B3%7D)
About the line y=8
![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}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B64x%20-%5Cfrac%7B32%7D%7B3%7Dx%5E3%20%2B%5Cfrac%7B4%7D%7B5%7Dx%5E5%5D%20%5CBig%7C_0%5E2%20%3D%5Cpi%20%2A%28128-%5Cfrac%7B256%7D%7B3%7D%20%2B%5Cfrac%7B128%7D%7B5%7D%29%3D%20%5Cfrac%7B1024%20%5Cpi%7D%7B5%7D)
About the line x=2
![V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5B%5Cfrac%7By%5E2%7D%7B2%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%2864%29%3D%2016%5Cpi)
Step-by-step explanation:
For this case we have the following functions:

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:

And we can find the volume with this formula:


![V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}](https://tex.z-dn.net/?f=%20V%20%3D%204%5Cpi%20%5B%5Cfrac%7Bx%5E5%7D%7B5%7D%5D%20%5CBig%7C_0%5E2%20%3D4%5Cpi%20%2A%5Cfrac%7B32%7D%7B5%7D%3D%20%5Cfrac%7B128%20%5Cpi%7D%7B5%7D)
About the y axis
For this case we need to find the function in terms of x like this:

but on this case we are just interested on the + part
as we can see on the second figure attached.
We can find the area like this:

And we can find the volume with this formula:


![V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B4y%20-y%5E2%20%2B%5Cfrac%7By%5E3%7D%7B12%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cpi%20%2A%5Cfrac%7B32%7D%7B3%7D%3D%20%5Cfrac%7B32%20%5Cpi%7D%7B3%7D)
About the line y=8
The figure 3 attached show the radius. We can find the area like this:

And we can find the volume with this formula:


![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}](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cpi%20%5B64x%20-%5Cfrac%7B32%7D%7B3%7Dx%5E3%20%2B%5Cfrac%7B4%7D%7B5%7Dx%5E5%5D%20%5CBig%7C_0%5E2%20%3D%5Cpi%20%2A%28128-%5Cfrac%7B256%7D%7B3%7D%20%2B%5Cfrac%7B128%7D%7B5%7D%29%3D%20%5Cfrac%7B1024%20%5Cpi%7D%7B5%7D)
About the line x=2
The figure 4 attached show the radius. We can find the area like this:

And we can find the volume with this formula:


![V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi](https://tex.z-dn.net/?f=%20V%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5B%5Cfrac%7By%5E2%7D%7B2%7D%5D%20%5CBig%7C_0%5E8%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%2864%29%3D%2016%5Cpi)
This one is simple since we already have the two x variables.equal. All we have to do is subtract the equations from one another to get the answer.
So i will subtract the left side by the other left side and the right side by the other right side
-8x - 8y -(-8x + 2y) = 0 -(-20)
distribute negative sign
-8x - 8y + 8x - 2y = 0 + 20
do the math
- 10y = 20
Y = -2
plug t into an equation
-8x -8 (-2) = 0
-8x + 16 = 0
-8x = -16
x = 2
answer (2, -2)
Answer: x = 63
Step-by-step explanation:
15/3 = 5
3 * 21 = 63
63/15 = 21/5