Question

EXAMPLE 2 The arc of the parabola y = 3x2 from (5, 75) to (10, 300) is rotated about the y-axis. Find the area of the resulting surface. SOLUTION 1 Using y = 3x2 and dy dx = we have, from this formula, S = 2πx ds = 10 2πx 1 + dy dx 2 dx 5 = 2π 10 x 1 + 36x2 dx 5 . Substituting u = 1 + 36x2, we have du = dx. Remembering to change the limits of integration, we have S = π 36 3601 u du 901 = π 36 3601 901 = . SOLUTION 2 Using x = y 3 and dx dy = we have S = 2πx ds = 300 2πx 1 + dx dy 2 dy 75 = 2π 300 y 3 1 + 1 12y dy 75 = π 3 300 12y + 1 dy 75 = π 36 3601 u du 901 (where u = 1 + 12y) = (as in Solution 1)

Answer 1

The area is given by the integral

`\displaystyle2\pi\int_5^(10)x√(1+(y'(x))^2)\,\mathrm dx=2\pi\int_5^(10)x√(1+36x^2)\,\mathrm dx`

Let
`u=1+36x^2\implies\mathrm du=72x\,\mathrm dx`, then the area is

`\displaystyle(2\pi)/(72)\int_(1+36\cdot5^2)^(1+36\cdot10^2)\sqrt u\,\mathrm du=\frac\pi{36}\int_(901)^(3601)u^(1/2)\,\mathrm du`

Since
`\left(\frac23u^(3/2)\right)'=u^(1/2)`, the integral is equal to

`(2\pi)/(3\cdot36)u^(3/2)\bigg|_(901)^(3601)=\boxed{(\left(3601^(3/2)-901^(3/2)\right)\pi)/(54)}`