## Wednesday, February 07, 2007

### 6.2 Volumes

In 6.1, we learned about the area between two curves, but now, Mr. French and the James Stewart have decided to make it more interesting by introducing a third dimension. We are now figuring out the volume between two curves, or the volume caused by the rotation of the area between two curves.

The general formula for finding the area between two curves is:
So let's say that we want to find the volume of a circle, using squares. The squares would come out the circle, with the base edge bounded by the circle. It would look something like this:

So, to calculate the volume we would use an infinite amount of infinitely thin boxes to calculate it, like a Riemann sum. Because it's a square we know that:And the formula for a circle is: where R is the radius. So the set up would look like this:

The green side, S, is double what the y value is for that x value, which means thatUsing this information, we can now define A in terms of x, like in the general equation for the volume, and find the volume in terms of R, which is an undefined constant.

Using the equation for the circle, we get that:And according to the general equation for volume:
So,Since the circle is also symmetrical about the y-axis, taking the integral from -R to R is the same thing as doubling the integral of 0 to R.Now, we just do what we know how to do - evaluate the integral, keeping in mind that R is a constant.And we're done!

There's another type of volume that Mr. French and James Stewart might ask you to find, and that's a rotation volume. It's when you take an area and rotate it around an axis, creating a solid with circular cross-sections. When the base of the area touches the rotation axis, it creates a disc volume, and when it doesn't, it creates a washer volume.

Take for instance the graph of the square root of x from 2 to 9.

And let's rotate it around the x-axis. And since the base of the area touches the axis of rotation, we have a disc volume, and because the cross sections are circles, the integral for the volume would be:
And the radius of the cross-section circle differs with the function. It is defined by the y-value, so R=Y (at x=9 the radius is 3, the same as the y-value of the function at 9). Since we know that y equals the square root of x, we can substitute the square root of x in for R. This gives us:

And from there, we all know how to solve this problem.

If we had rotated the function around the line x=-3, then we would have had a washer volume, since the base of the function wouldn't have touched the rotation axis. To solve that problem, we would have had to find the volume without the space, and then subtracted the space. There would be two radii, one representing the radius of the empty cylinder in the middle and one representing the radius of the whole volume.

To solve this problem, we would use:
In this case, r would be a constant of 3, while R would be 3+ y-value of the function or
So we can plug all of this into the big equation and get:

And we all know how to solve the problem from there!

If you need any more help on this concept, you should check out these websites:
And finally I just wanted to remind everyone that our favorite bunch of castaways are back on tonight after a long winter hiatus.

Just as a recap, Jack, Kate and Sawyer captured by the Others, and Jack had Benry on the table in the middle of back surgery, when he insisted that Kate and Sawyer be let go or he'd leave Benry on the table to die. So that should make for an interesting episode tonight!
Also, reminder to Lauren for the next post.