Okay, so it appears I spoke too soon last time when I said I understood this concept. I clearly do not, as I am struggling again. Problem 43 from Section 6.1 of my text book says:
I began by drawing this graph. I want to rotate the region shown in blue around the line x=3. I figured the easiest way to do so would be to calculate the entire volume of the rectanglar prism (that is, all three colored areas put together) and then subtract the purple and blue regions. I also know that my function is y=4-x2. That means that x2 = 4 - y. For the green region, y goes from 0 to 4, so those are the bounds of my integral. The radius of the solid is the width of the 2D shape. So what I ended up with was this:
\(\displaystyle \pi \int _0^5\:3^2dy-\pi \int _0^5\:1^2dy-\pi \int _0^4\left(2-\sqrt{4-y}\right)^2dy\)
And after evaluating that integral, I come up with 112pi/3, which is not even close to the expected answer of 16pi.
I began by drawing this graph. I want to rotate the region shown in blue around the line x=3. I figured the easiest way to do so would be to calculate the entire volume of the rectanglar prism (that is, all three colored areas put together) and then subtract the purple and blue regions. I also know that my function is y=4-x2. That means that x2 = 4 - y. For the green region, y goes from 0 to 4, so those are the bounds of my integral. The radius of the solid is the width of the 2D shape. So what I ended up with was this:
\(\displaystyle \pi \int _0^5\:3^2dy-\pi \int _0^5\:1^2dy-\pi \int _0^4\left(2-\sqrt{4-y}\right)^2dy\)
And after evaluating that integral, I come up with 112pi/3, which is not even close to the expected answer of 16pi.
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