How do you setup an integral to compute volume of revolution obtained by rotating the region bounded by curves y=x^2,x=1 and x=2 and the x-axis around the axis y=-1.
Again, someone check me on this. we will use the disk method.
recall, by the disk method, V = int{Area}dx = int{pi*r^2}dx = int{pi*r*f(x)}dx
so we have the radius from the x-axis being x^2, but we are rotating about a line 1 unit from the x-axis, so the radius we will use is 1 + x^2 . the limits as you can see from the diagram are between x= 1 and x = 2. so here goes.
V = int{pi(1 + x^2)^2}dx evaluated between 1 and 2
=> V = pi*int{(1 + x^2)^2}dx
=> V = pi*int{1 + 2x^2 + x^4}dx
=> V = pi[x + (2/3)x^3 + (1/5)x^5] between 1 and 2
=> V = pi[(2 + (2/3)2^3 + (1/5)2^5) - (1 + 2/3 + 1/5)]
=> V = 178pi/15