In order to calculate this volume we are in some sense going to cut it into (infinitely slim) slices.
We envision the region, to help us with this, I have enclosed the graph where the region is the part beneath the curve. We note that y=x^2-1y=x2−1 crosses the line x=5x=5 where y=24y=24 and that it crosses the line y=0y=0 where x=1x=1 graph{x^2-1 [1, 5, -1, 24]}
When cutting this region in horizontal slices with height dydy (a very small height). The length of these slices depends very much on the y coordinate. to calculate this length we need to know the distance from a point (y,x)(y,x) on the line y=x^2-1y=x2−1 to the point (5,y). Of course this is 5-x5−x, but we want to know how it depends on yy. Since y=x^2-1y=x2−1, we know x^2=y+1x2=y+1, since we have x>0x>0 for the region we are interestend in, x=sqrt(y+1)x=√y+1, therefore this distance dependant on yy, which we shall denote as r(y)r(y) is given by r(y)=5-sqrt(y+1)r(y)=5−√y+1.
Now we rotate this region around x=5x=5, this means that every slice becomes a cylinder with height dydy and radius r(y)r(y), therefore a volume pir(y)^2dyπr(y)2dy. All we need to do now is add up these infinitely small volumes using integration. We note that yy goes from 00 to 2424.
V=int_0^24pir(y)^2dy=piint_0^24(5-sqrt(y+1))^2dy=piint_0^24 (25-10sqrt(y-1)+y+1)dy=piint_0^24 (26-10sqrt(y+1)+y)dy=pi[26y-20/3(y+1)^(3/2)+y^2/2]_0^24=pi(26*24-20/3(25)^(3/2)+20/3+24^2/2)=pi(85+1/3).
