Question

Find the surface area of the solid of revolution obtained by rotating the curve x=(2y−y^2)^(1/2) from y=8/16 to y=31/16 about the y-axis?

Answer 1

`color(blue)[S_A=2piint_(8/16)^(31/16)(2y−y^2)^(1/2)*sqrt(1+((2-2y)^2)/((2y−y^2)(4)))*dy=(46pi)/16]`

Explanation:

The surface area of solid of revolution around `"y-axis"` is given by:

`color(red)[S_A=2piint_a^bx*sqrt(1+(x')^2)*dy`

The interval of integral is `y in [8/16,31/16]`

now lets setup the integral of surface area

`S_A=2piint_(8/16)^(31/16)((2y−y^2)^(1/2))*sqrt(1+(1/2(2y-y^2)^(-1/2)(2-2y))^2)*dy`

`S_A=2piint_(8/16)^(31/16)(2y−y^2)^(1/2)*sqrt(1+((2-2y)^2)/((2y−y^2)(4)))*dy`

After simplified it:

`=2piint_(8/16)^(31/16)[(i*sqrt(-(y-2)*y))/sqrt((y-2)*y)]*dy`

`=2piint_(8/16)^(31/16)i*sqrt(-1)*dy=2piint_(8/16)^(31/16)i^2*dy`

`=2piint_(8/16)^(31/16)1*dy=2pi[y]_(8/16)^(31/16)`

`=2pi[(31/16)-(8/16)]=(46pi)/16`

Note that

`i^2=1`