Question

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?

Answer 1

Take a piece of the curve whose length is the differential of arc length `ds = sqrt(1+(f'(x))^2) dx`

The surface area of the corresponding piece of the solid is

the circumference of revolution time the differential of arc length

which is

`2pif(x)sqrt(1+(f'(x))^2) dx`

Using just the values `0 <= x <=a`, the surface area of the solid is

`2piint_0^a f(x)sqrt(1+(f'(x))^2) dx = 2pi int_0^a sqrt(r^2-x^2)sqrt(1+((-x)/sqrt(r^2-x^2))^2) dx`

`= 2pi int_0^a r dx`

`= 2pi r a`

Answer 2

`A=2piar`

Explanation:

The formular is
`A=2piint_a^bf(x)sqrt(1+(f'(x))^2)dx`
`f(x)=y=sqrt(r^2-x^2)`
`f'(x)=-(x)/(sqrt(r^2-x^2)`
`A=2piint_0^asqrt(r^2-x^2)sqrt(1+(x^2)/(r^2-x^2))dx`
`<=>2piint_0^asqrt(r^2-x^2)sqrt(r^2/(r^2-x^2))dx`
`<=>2piint_0^ardx`
`<=>2pi[xr]_0^a`
`<=>2piar`