Find the volume of the solid that results where the shaded region is revolved about the y-axis.

y=3-2x

I'm not sure if you solve for x then integrate or if you just integrate the equation as it is.(Wondering)(Thinking)(Nerd)

Printable View

- Mar 3rd 2009, 02:27 PMkinana18Volumes by Rotation about y-axis
Find the volume of the solid that results where the shaded region is revolved about the y-axis.

y=3-2x

I'm not sure if you solve for x then integrate or if you just integrate the equation as it is.(Wondering)(Thinking)(Nerd) - Mar 3rd 2009, 02:55 PMAmanda H
You want to rearrange your equation so you have $\displaystyle x = ... $. I'm assuming you can do that. Then integrate like this:

$\displaystyle \int \pi x^2 .dy $ You can take $\displaystyle \pi $ outside the integral

$\displaystyle \pi \int x^2 .dy $

Here you substitute in your rearranged equation, square it and then integrate between the intervals that you'll probably have to work out (I'm not sure since you've made no mention of them in the question). - Mar 3rd 2009, 03:02 PMSoroban
Hello, kinana18!

Quote:

Find the volume of the solid that results where the shaded region . Where?

is revolved about the y-axis. . $\displaystyle y\:=\:3-2x$

I'm not sure if you solve for x then integrate

or if you just integrate the equation as it is.

I have no idea what you're talking about . . . do you?

The only region that makes any sense looks like this:Code:`|`

3 *

|\

|:\

|::\

|:::\

|::::\

|:::::\

|::::::\

- + - - - * - -

| 1½

We have a right circular cone with $\displaystyle r = \tfrac{3}{2},\;h=3$

Its volume is: .$\displaystyle V \:=\:\tfrac{1}{3}\pi r^2h \:=\:\tfrac{1}{3}\pi\left(\tfrac{3}{2}\right)^2(3) \:=\:\frac{9\pi}{4}$

If we**must**use Calculus, I suggest Cylindrical Shells.

The formula is: .$\displaystyle V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx$

And we have: . $\displaystyle V \;=\;2\pi\int^{\frac{3}{2}}_0x(3-2x)\,dx $ . . . . etc.