Answer to Question #332064 in Calculus for gel

Question

Find the area of the surface generated by revolving the curve y=√25-x², from x = - 2 to x = 3 , about the x-axis.

Answer 1

$A = 2 \pi \int_a^by \cdot \sqrt{1+(y')^2}$ dx

$y = \sqrt{25-x^2}, y' = -\cfrac{x}{\sqrt{25-x^2}}$

$A = 2 \pi \int_{-2}^3 (\sqrt{25-x^2} \cdot\sqrt{1 + \cfrac{x^2}{25-x^2}})dx =\\ =2\pi \int_{-2}^3(\sqrt{25-x^2} \cdot \sqrt{\cfrac{25}{25-x^2}})dx = 2\pi \int_{-2}^3 5 dx = 2\pi \cdot5x|_{-2}^3 = \\ =2\pi (15+10) = 50\pi$