How do you find the surface area generated when the curve y=sqrt(r^2-x^2), 0<=x<=a is rotated about the x axis?

Apr 17, 2018

Take a piece of the curve whose length is the differential of arc length $\mathrm{ds} = \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

The surface area of the corresponding piece of the solid is

the circumference of revolution time the differential of arc length

which is

$2 \pi f \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

Using just the values $0 \le x \le a$, the surface area of the solid is

$2 \pi {\int}_{0}^{a} f \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx} = 2 \pi {\int}_{0}^{a} \sqrt{{r}^{2} - {x}^{2}} \sqrt{1 + {\left(\frac{- x}{\sqrt{{r}^{2} - {x}^{2}}}\right)}^{2}} \mathrm{dx}$

$= 2 \pi {\int}_{0}^{a} r \mathrm{dx}$

$= 2 \pi r a$

Apr 17, 2018

$A = 2 \pi a r$

Explanation:

The formular is
$A = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$
$f \left(x\right) = y = \sqrt{{r}^{2} - {x}^{2}}$
f'(x)=-(x)/(sqrt(r^2-x^2)
$A = 2 \pi {\int}_{0}^{a} \sqrt{{r}^{2} - {x}^{2}} \sqrt{1 + \frac{{x}^{2}}{{r}^{2} - {x}^{2}}} \mathrm{dx}$
$\iff 2 \pi {\int}_{0}^{a} \sqrt{{r}^{2} - {x}^{2}} \sqrt{{r}^{2} / \left({r}^{2} - {x}^{2}\right)} \mathrm{dx}$
$\iff 2 \pi {\int}_{0}^{a} r \mathrm{dx}$
$\iff 2 \pi {\left[x r\right]}_{0}^{a}$
$\iff 2 \pi a r$