Answer:
volume V of the solid

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

for k=1,2,..., n
We then add up the volumes of all these slices

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

we also know that

and

so we have, after replacing and simplifying, the sum of the slices equals

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

and the volume V of our solid is
