1. ## Volumes of Revolution

find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3
Using the Disk Method

I used the shell method and got: 175pi/3

But I need to know how to set up the disk integral

Also,

Y=X^2+1, X=0, Y=5 revolved around Y=7

using the shell method i think i got 12pi, but that may be wrong, and I need to know how to set up the disk method.

thanks so much

2. ## Volume of revolution

Hello terrytriangle
Originally Posted by terrytriangle
find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3
Please clarify this question. The curve $y = x^2 +1$, the line $y =0$ (i.e. the $x$-axis) and the line $x = 2$ does not enclose an area. Do you mean the area enclosed by

• $y=x^2 +1, y=0, x=2$ and $x=3$?
• Or $y=x^2+1, x=0, y=0$ and $x=2$?
• Or something else?

3. Originally Posted by terrytriangle
find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3
Using the Disk Method

I used the shell method and got: 175pi/3

...
1. If I understand your question correctly the solid looks like a kind of funnel (upside down). (See attachment)

2. The solid consists of two different kinds of solids:
The base is a cylindrical ring with height of 1.
The top is the funnel mentioned above.

3. The base area of one disk is

$b = \pi((3-x)^2-1^2)=(x^2-6x+8)\cdot \pi$

the height of one disk is

$h = dy$ From

$\dfrac{dy}{dx} = 2x~\implies~dy=2x\cdot dx$

4. The complete volume is:

$V = \pi(3^2-1^2)\cdot 1+\int_0^2\left(\pi((x-3)^2-1^2)\cdot 2x \right)dx$

I leave the rest for you.

By the way: Your result must be too large: The solid can be placed completely into a cylinder with a base diameter of 6 and a height of 5. Such a cylinder has a volume of $45\pi$

4. thank you i figured it out now

5. earboth, terrytriangle said, "find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3", which, as Granddad said, doesn't quite make sense. You have assumed the region enclosed by $y= x^2+ 1$, y= 0, x= 2, x= 3, revolved around x= 0.

6. Originally Posted by HallsofIvy
earboth, terrytriangle said, "find the volume of the area enclosed by Y=X^2+1, Y=0, X=2 and revolved around X=3", which, as Granddad said, doesn't quite make sense. You have assumed the region enclosed by $y= x^2+ 1$, y= 0, x= 2, x= 3, revolved around x= 0.