# Find the volume Calculus 2

• Jun 7th 2007, 04:04 PM
Whitewolfblue
Find the volume Calculus 2
I am really lost.

1.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=8 x^2, x = 1, y = 0, about the x-axis

2.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2, y = 0, x = 3, about the y-axis

3.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = 28 x - 4 x^2, y = 0;

Kiba
• Jun 7th 2007, 04:20 PM
qbkr21
Re:
1.

64pi/5

2.

81pi/2

3.

4802pi/3
• Jun 7th 2007, 04:21 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
I am really lost.

1.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=8 x^2, x = 1, y = 0, about the x-axis

See the graph below. our limits of integration are $\displaystyle x = 0 \mbox { and } x = 1$

We proceed by the disk method.

Recall that by the disk method, we find the volume by integrating the area of disks that we slice the function into. thus:

$\displaystyle V = \int_{a}^{b} \pi r^2 dx = \int_{a}^{b} \pi f(x)^2 dx$, when we are rotating about the x-axis and where $\displaystyle a \mbox { and } b$ are our limtis of integration

$\displaystyle V = \int_{0}^{1} \pi \left( 8x^2 \right)^2 dx$

$\displaystyle = 64 \pi \int_{0}^{1}x^4 dx$

$\displaystyle = 64 \pi \left[ \frac {1}{5}x^5 \right]_{0}^{1}$

$\displaystyle = \frac {64 \pi}{5}$

Quote:

2.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2, y = 0, x = 3, about the y-axis

3.) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = 28 x - 4 x^2, y = 0;

Kiba
the two above are done similar. except since we are rotating about the y-axis, we have to change the functions to functions of y, and the limits in terms of y if we want to use the disk method. or we can leave everything in terms of x and use the shell method.

you may want to do a search on "volume" to see other threads where more explanations are given

EDIT: well, it seems qbkr21 is providing you with the answers, so now all you have to do is show how he got to those answers. tell us if you have any problems

EDIT: Here are some threads you might want to check out

http://www.mathhelpforum.com/math-he...nd-volume.html

http://www.mathhelpforum.com/math-he...ergrating.html

http://www.mathhelpforum.com/math-he...-rotation.html

http://www.mathhelpforum.com/math-he...ng-volume.html

http://www.mathhelpforum.com/math-he...ng-volume.html
• Jun 7th 2007, 06:05 PM
Whitewolfblue
So far number two is incorrect, so I am working it out now.

I need to just change the function in terms of "y" correct?

But how do I change the limits?

Kiba

basically when I look at it y = x^2 correct? So then the square root of Y would equal x. If y=0 then the square root of zero is zero. This graph isn't adding up to give me a valid area "below the curve" so to speak right? Or am I looking at this incorrectly?
• Jun 7th 2007, 06:06 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
So far number two is incorrect, so I am working it out now.

I need to just change the function in terms of "y" correct?

But how do I change the limits?

Kiba

use the formulas. plug in the limits for x and solve for y
• Jun 7th 2007, 06:30 PM
qbkr21
Re:
Re: