Application of integrals

x^2/9 + y^2 = 1
y^2 = 1 - x^2/9
y = +/- sqrt(1 - x^2/9)
So the upper part of the curve is y = sqrt(1 - x^2/9) = t and the lower part is y = - sqrt(1 - x^2/9) = - t
Volume generated when rotates about line y = 5
= ∫ π[5 - ( - t)]^2 dx - ∫ π[5 - t]^2 dx from x = - 3 to x = 3.
= π ∫ [(5 + t)^2 - (5 - t)^2] dx
= π ∫(5 + t + 5 - t)(5 + t - 5 + t) dx
= π ∫(10)(2t) dx
= 20π ∫ sqrt(1 - x^2/9) dx
= 20π/3 ∫ sqrt(9 - x^2) dx from x = - 3 to x = 3
= 40π/3 ∫ sqrt(9 - x^2) dx from x = 0 to x = 3 since this is an even function.
Using substitution x = 3 sin z
Changing limits to from 0 to π/2 we get
Volume = 40π/3(9π/4) = 30π^2.