# volumes of solids of revolution

Submitted by Jeanvill Palada... on March 19, 2016 - 4:54pm

problem no.1

find the volume of the solid that results when the region enclosed by y=squareroot of x,y=0, and x=9 is revolved about the line x=9

prob.no.2

find the volume of the solid that results when the region in problem no. 1 is revolved about the line y=3

- Add new comment
- 1123 reads

## Re: volumes of solids of revolution

Volume generated when the area is rotated about the line x = 9$V = 2\pi {\displaystyle \int_a^b} R \, dA$

$V = 2\pi {\displaystyle \int_{x_1}^{x_2}} (9 - x)(y \, dx)$

$V = 2\pi {\displaystyle \int_0^9} (9 - x)(\sqrt{x} \, dx)$

$V = 129.6\pi ~ \text{unit}^3$

Volume generated when the area is rotated about the line y = 3$V = \pi {\displaystyle \int_{x_1}^{x_2}} ({R_{\text{outer}}}^2 - {R_{\text{inner}}}^2) \, dx$

$V = \pi {\displaystyle \int_0^9} \left[ 3^2 - (3 - y)^2 \right] \, dx$

$V = \pi {\displaystyle \int_0^9} \left[ 9 - (3 - \sqrt{x})^2 \right] \, dx$

$V = 67.5\pi ~ \text{unit}^3$

## Re: volumes of solids of revolution

Thank you po

## Re: volumes of solids of revolution

Solution by Pappus's Theorem$A = \frac{2}{3}(3)(9) = 18 ~ \text{unit}^2$

$\bar{x} = \frac{2}{5}(9) = 3.6 ~ \text{units}$

$\bar{y} = \frac{3}{8}(3) = 1.125 ~ \text{units}$

Volume generated when the area is rotated about the line x = 9$V = 18 \times 2\pi (3.6)$

$V = 129.6\pi ~ \text{unit}^3$

Volume generated when the area is rotated about the line y = 3$V = 18 \times 2\pi (3 - 1.125)$

$V = 67.5\pi ~ \text{unit}^3$

## Re: volumes of solids of revolution

pasagot nga po ulit ito..thanks

## Add new comment