# 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?

May 25, 2018

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 $\text{y-axis}$ is given by:

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

The interval of integral is $y \in \left[\frac{8}{16} , \frac{31}{16}\right]$

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:

$= 2 \pi {\int}_{\frac{8}{16}}^{\frac{31}{16}} \left[\frac{i \cdot \sqrt{- \left(y - 2\right) \cdot y}}{\sqrt{\left(y - 2\right) \cdot y}}\right] \cdot \mathrm{dy}$

$= 2 \pi {\int}_{\frac{8}{16}}^{\frac{31}{16}} i \cdot \sqrt{- 1} \cdot \mathrm{dy} = 2 \pi {\int}_{\frac{8}{16}}^{\frac{31}{16}} {i}^{2} \cdot \mathrm{dy}$

$= 2 \pi {\int}_{\frac{8}{16}}^{\frac{31}{16}} 1 \cdot \mathrm{dy} = 2 \pi {\left[y\right]}_{\frac{8}{16}}^{\frac{31}{16}}$

$= 2 \pi \left[\left(\frac{31}{16}\right) - \left(\frac{8}{16}\right)\right] = \frac{46 \pi}{16}$

Note that

${i}^{2} = 1$