Answer to Question #186710 in Calculus for Jethro

Question

Determine the surface area of the solid obtained by rotating:

1) x² + y² = 16 from x = 2 to x = 4

2) y² = 12x from x = 0 to x = 3

3) y = x³ from x = 0 to x = 1

Answer 1

1) x² + y² = 16 from x = 2 to x = 4

$f(x)==\sqrt{16-y^2}$

The surface area of the curve is given by$S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x$

First, find the derivative

$f '\left(x\right)=\left(\sqrt{16 - x^{2}}\right)'=- \frac{x}{\sqrt{16 - x^{2}}}$

Finally, calculate the integral

$S = \int_{2}^{4} 2 \pi \sqrt{16 - x^{2}} \sqrt{\left(- \frac{x}{\sqrt{16 - x^{2}}}\right)^{2} + 1} d x$

$=\int_{2}^{4} 8 \pi \sqrt{- \frac{1}{x^{2} - 16}} \sqrt{16 - x^{2}} d x$

=$\int_{2}^{4}\left( 8 \pi \sqrt{- \frac{1}{x^{2} - 16}} \sqrt{16 - x^{2}} \right)dx=\int_{2}^{4}\left( 8 \pi \right)dx$

$=\int_{2}^{4}\left( 8 \pi \right)dx=\left(8 \pi x\right)|_{\left(x=4\right)}-\left(8 \pi x\right)|_{\left(x=2\right)}=16 \pi$

$=16 \pi\\ \approx \boxed{50.2654824574367}$

2) y² = 12x from x = 0 to x = 3

$f\left(x\right)=2 \sqrt{3} \sqrt{x}$

he surface area of the curve is given by$S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x$

$f '\left(x\right)=\left(2 \sqrt{3} \sqrt{x}\right)'=\frac{\sqrt{3}}{\sqrt{x}}$

calculate the integral

$S = \int_{0}^{3} 2 \pi 2 \sqrt{3} \sqrt{x} \sqrt{\left(\frac{\sqrt{3}}{\sqrt{x}}\right)^{2} + 1} d x \\=\int_{0}^{3} 4 \sqrt{3} \pi \sqrt{x} \sqrt{\frac{x + 3}{x}} d x$

$=\int_{0}^{3}\left( 4 \pi \sqrt{3 x + 9} \right)dx$

$\int_{0}^{3}\left( 4 \pi \sqrt{3 x + 9} \right)dx=\left(\frac{8 \sqrt{3} \pi \left(x + 3\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=3\right)}-\left(\frac{8 \sqrt{3} \pi \left(x + 3\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=0\right)}=- 24 \pi + 48 \sqrt{2} \pi$

$=- 24 \pi + 48 \sqrt{2} \pi\\\approx \boxed{137.860157345447}$

3) y = x³ from x = 0 to x = 1

f(x)=x³

$S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x$

$f '\left(x\right)=\left(x^{3}\right)'=3 x^{2}$

$S = \int_{0}^{1} 2 \pi x^{3} \sqrt{\left(3 x^{2}\right)^{2} + 1} d x=\int_{0}^{1} 2 \pi x^{3} \sqrt{9 x^{4} + 1} d x$

$\int_{0}^{1}\left( 2 \pi x^{3} \sqrt{9 x^{4} + 1} \right)dx \\=\left(\frac{\pi \left(9 x^{4} + 1\right)^{\frac{3}{2}}}{27}\right)|_{\left(x=1\right)}-\left(\frac{\pi \left(9 x^{4} + 1\right)^{\frac{3}{2}}}{27}\right)|_{\left(x=0\right)}\\=- \frac{\pi}{27} + \frac{10 \sqrt{10} \pi}{27}$

$\approx \boxed{3.56312185201375}$