Calculus 2 Volume
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Dr.Peterson
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Please show us what you've tried, so we can see what sort of help you need.
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The hardest part is usually to keep track of where the axis is, in order to get the right radii. They will be the difference between x-coordinates on the appropriate part of each curve and on the axis.
I don't think I'd prefer the washer method for this one!
Dr.Peterson
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You solved both equations for x, but forgot the \(\pm\). And in your integral, you didn't make r1 and r2 functions of y, and just used the maximum value of each.
You'll need two separate integrals, one from y=1 to 2, and another from y=2 to 3. Each will take r1 at the right side and r2 at the left side of the same curve.
HallsofIvy
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Notice that the "washer method" is really two "disk" methods:
\(\displaystyle \pi \int_a^b (f^2- g^2)x= \pi\int_a^b f^2dx- \pi\int_a^b g^2dx\).
First find the volume when \(\displaystyle y= 3- x^2\), for x between -1 and 1, rotated around y= -1, \(\displaystyle \pi\int_{-1}^1 (3- x^2- (-1))^2 dx= \pi\int_{-1}^1 (4- x^2)^2 dx\).
Then find the volume when \(\displaystyle y= x^2+ 2\), for x between -1 and 1, rotated about y= -1, \(\displaystyle \pi\int_{-1}^1 (x^2+ 2- (-1))^2 dx= \pi\int_{-1}^1 (x^2+ 3)^2 dx\).
And subtract!
\(\displaystyle \pi \int_a^b (f^2- g^2)x= \pi\int_a^b f^2dx- \pi\int_a^b g^2dx\).
First find the volume when \(\displaystyle y= 3- x^2\), for x between -1 and 1, rotated around y= -1, \(\displaystyle \pi\int_{-1}^1 (3- x^2- (-1))^2 dx= \pi\int_{-1}^1 (4- x^2)^2 dx\).
Then find the volume when \(\displaystyle y= x^2+ 2\), for x between -1 and 1, rotated about y= -1, \(\displaystyle \pi\int_{-1}^1 (x^2+ 2- (-1))^2 dx= \pi\int_{-1}^1 (x^2+ 3)^2 dx\).
And subtract!
Dr.Peterson
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I don't see that you've done what I said. Can you explain each part of this?
What are your two radii, and where do they show up on the graph?