Find volume of resulting solid using any method
The picture below illustrates the problem.
From this picture we can see that the tip of the sideways parabola is to be rotated around the axis y=5 that it is lying upon. When y=6, x=0 (vertex) and the part of the parabola we need is therefore between x=0 and 1.
Since x=(y-6)^2, then y=6±√x. This gives us two circles for each x value. The larger circle has a radius of 6+√x-5=1+√x and the smaller circle has a radius of 1-√x. The difference between the areas of these two circles gives the area of a "washer": π((1+√x)^2-(1-√x)^2)=π(1+√x-1+√x)(1+√x+1-√x)=4π√x. The volume of the washer is 4π√xdx for an infinitesimally thin washer, thickness dx. If we integrate 4π∫(√xdx)=4π[(2/3)x^(3/2)] for 0≤x≤1, we get 8π/3=8.3776 cu units = volume of revolution.
x=(y-6)^2 so √x=y-6 and dx=2(y-6)dy. The volume of the washer is 4π(y-6)dx=8π∫((y-6)^2dy) = 8π∫((y^2-12y+36)dy). The limits are found through solving x=1=(y-6)^2 making the intersections (1,5) and (1,7). But when x=0, y=6, so the limits are 6≤y≤7 not 5≤y≤7.
Integrating: 8π[y^3/3-6y^2+36y] for 6≤y≤7=8π((7^3-6^3)/3-6(7^2-6^2)+36(7-6))=8π(127/3-78+36)=8π/3.
The same result applies if we take the limits 5≤y≤6=8π((6^3-5^3)/3-6(6^2-5^2)+36(6-5))=8π(91/3-66+36)=8π/3.
Had we used 5≤y≤7 we would have included the same volume twice, so the result would have been 16π/3. Read More: ...