Answer:
volume V of the solid
![\boxed{V=\displaystyle\frac{125\pi}{12}}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B12%7D%7D)
Step-by-step explanation:
The situation is depicted in the picture attached
(see picture)
First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each
[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]
Now, we slice our solid into n slices.
Each slice is a quarter of cylinder 5/n thick and has a radius of
-k(5/n) + 5 for each k = 1,2,..., n (see picture)
So the volume of each slice is
![\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%5Cpi%28-k%285%2Fn%29%20%2B%205%20%29%5E2%2A%285%2Fn%29%7D%7B4%7D)
for k=1,2,..., n
We then add up the volumes of all these slices
![\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B%5Cpi%28-%285%2Fn%29%20%2B%205%20%29%5E2%2A%285%2Fn%29%7D%7B4%7D%2B%5Cdisplaystyle%5Cfrac%7B%5Cpi%28-2%285%2Fn%29%20%2B%205%20%29%5E2%2A%285%2Fn%29%7D%7B4%7D%2B...%2B%5Cdisplaystyle%5Cfrac%7B%5Cpi%28-n%285%2Fn%29%20%2B%205%20%29%5E2%2A%285%2Fn%29%7D%7B4%7D)
Notice that the last term of the sum vanishes. After making up the expression a little, we get
![\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cleft%5B%28-%285%2Fn%29%2B5%29%5E2%2B%28-2%285%2Fn%29%2B5%29%5E2%2B...%2B%28-%28n-1%29%285%2Fn%29%2B5%29%5E2%5Cright%5D%3D%5C%5C%5C%5C%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D%28-k%285%2Fn%29%2B5%29%5E2)
But
![\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D%28-k%285%2Fn%29%2B5%29%5E2%3D%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D%28%285%2Fn%29%5E2k%5E2-%2850%2Fn%29k%2B25%29%3D%5C%5C%5C%5C%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cleft%28%285%2Fn%29%5E2%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%5E2-%2850%2Fn%29%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%2B25%28n-1%29%5Cright%29)
we also know that
![\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%5E2%3D%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%282n-1%29%7D%7B6%7D)
and
![\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%3D%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%7D%7B2%7D)
so we have, after replacing and simplifying, the sum of the slices equals
![\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cleft%28%285%2Fn%29%5E2%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%5E2-%2850%2Fn%29%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7Dk%2B25%28n-1%29%5Cright%29%3D%5C%5C%5C%5C%3D%5Cdisplaystyle%5Cfrac%7B5%5Cpi%7D%7B4n%7D%5Cleft%28%5Cdisplaystyle%5Cfrac%7B25%7D%7Bn%5E2%7D.%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%282n-1%29%7D%7B6%7D-%5Cdisplaystyle%5Cfrac%7B50%7D%7Bn%7D.%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%2B25%28n-1%29%5Cright%29%3D%5C%5C%5C%5C%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B24%7D.%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%282n-1%29%7D%7Bn%5E3%7D)
Now we take the limit when n tends to infinite (the slices get thinner and thinner)
![\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B24%7D%5Cdisplaystyle%5Clim_%7Bn%20%5Crightarrow%20%5Cinfty%7D%5Cdisplaystyle%5Cfrac%7Bn%28n-1%29%282n-1%29%7D%7Bn%5E3%7D%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B24%7D%5Cdisplaystyle%5Clim_%7Bn%20%5Crightarrow%20%5Cinfty%7D%282-3%2Fn%2B1%2Fn%5E2%29%3D%5C%5C%5C%5C%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B24%7D.2%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B12%7D)
and the volume V of our solid is
![\boxed{V=\displaystyle\frac{125\pi}{12}}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D%5Cdisplaystyle%5Cfrac%7B125%5Cpi%7D%7B12%7D%7D)