# Thread: Volumes of solids by slicing help

1. ## Volumes of solids by slicing help

An equation of the curve in the figure is y^2 = x^3

Line 4, OAC is the shaded curve
Code:
     B  |               C(4,8)
8+ - - - - - -*
|           *|
|         *::|
|       * :::|
|     * :::::|
|  *:::::::::|
--*------------+
O  |              4  A
Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
revolved OAC about the line AC

Help the only problem i had is to subtract the radiuses to get to that given area of OAC...
ill handle the integration

Can you check my solution?

$\pi\int_{0}^{8} y^{\frac{2}{3}}$

$\pi\frac{3}{5}y^{\frac{5}{3}}\big|_{0}^{8}$

$\frac{96}{5}\pi$

An equation of the curve in the figure is y^2 = x^3

Line 4, OAC is the shaded curve
Code:
     B  |               C(4,8)
8+ - - - - - -*
|           *|
|         *::|
|       * :::|
|     * :::::|
|  *:::::::::|
--*------------+
O  |              4  A
Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
revolved OAC about the line AC

Help the only problem i had is to subtract the radiuses to get to that given area of OAC...
ill handle the integration

Can you check my solution?

$\pi\int_{0}^{8} y^{\frac{2}{3}}$

$\pi\frac{3}{5}y^{\frac{5}{3}}\big|_{0}^{8}$

$\frac{96}{5}\pi$
that's not correct i'm afraid.

you are using the disk method, so to rotate about the interval $[a,b]$ (whether with respect to $x$ or $y$), the formula for the Disk method is:

$V = \pi \int_{a}^{b} [R(x \mbox { or } y)]^2$

where $R(x \mbox { or } y)$ is the radius.

since you're doing $V = \pi \int_{a}^{b} [R( {\color {red} y})]^2~{\color {red} dy}$, where $[a,b]$ is the interval on the y-axis, with $b > a$

your radius here is: $R(y) = \left(4 - y^{2/3} \right)$