use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 )​, (15 comma 0 )​, and (0 comma 15 )and whose cross sections perpendicular to the base and parallel to the​ y-axis are semicircles

Answer:volume V of the solid [tex]\boxed{V=\displaystyle\frac{125\pi}{12}}[/tex]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  [tex]\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}[/tex]for k=1,2,..., nWe then add up the volumes of all these slices[tex]\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}[/tex]Notice that the last term of the sum vanishes. After making up the expression a little, we get[tex]\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[/tex]But[tex]\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)[/tex]we also know that[tex]\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}[/tex]and[tex]\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}[/tex]so we have, after replacing and simplifying, the sum of the slices equals[tex]\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}[/tex]Now we take the limit when n tends to infinite (the slices get thinner and thinner)[tex]\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}[/tex]and the volume V of our solid is[tex]\boxed{V=\displaystyle\frac{125\pi}{12}}[/tex]