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^2-1# crosses the line #x=5# where #y=24# and that it crosses the line #y=0# where #x=1# graph{x^2-1 [1, 5, -1, 24]}
When cutting this region in horizontal slices with height #dy# (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)# on the line #y=x^2-1# to the point (5,y). Of course this is #5-x#, but we want to know how it depends on #y#. Since #y=x^2-1#, we know #x^2=y+1#, since we have #x>0# for the region we are interestend in, #x=sqrt(y+1)#, therefore this distance dependant on #y#, which we shall denote as #r(y)# is given by #r(y)=5-sqrt(y+1)#.
Now we rotate this region around #x=5#, this means that every slice becomes a cylinder with height #dy# and radius #r(y)#, therefore a volume #pir(y)^2dy#. All we need to do now is add up these infinitely small volumes using integration. We note that #y# goes from #0# to #24#.
#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)#.
