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

$\displaystyle2\pi\int_5^{10}x\sqrt{1+(y'(x))^2}\,\mathrm dx=2\pi\int_5^{10}x\sqrt{1+36x^2}\,\mathrm dx$

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

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

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

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