set up integral for volume of sqrt[4 - x^2], 0 <= x <=
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It would appear they want you to use the washer method. Shells, I think, would be a little easier.
Here's shells. It'll be in terms of x. You try setting up the washer method in terms of y. See you get the same thing.
\(\displaystyle \L\\2{\pi}\int_{0}^{2}(x+2)(\sqrt{4-x^{2}})dx\)
Here's shells. It'll be in terms of x. You try setting up the washer method in terms of y. See you get the same thing.
\(\displaystyle \L\\2{\pi}\int_{0}^{2}(x+2)(\sqrt{4-x^{2}})dx\)
Re: volume
Hello, mathhelp!
I was going to comment that the washer method is dangerous
. . near the top of the region.
Then I realized that the graph is drawn incorrectly.
The graph of: \(\displaystyle \,y\:=\:\sqrt{4\,-\,x^2}\,\) is a quarter-circle in Quadrant I.
The volume can be found either way:
Washers: \(\displaystyle \L\:V \:=\:\pi\int^{\;\;\;2}_0\left(\sqrt{4-y^2} \,+\,2\right)^2\,dy\)
. . Shells: \(\displaystyle \L\:V \:=\:2\pi\int^{\;\;\;2}_0 x\sqrt{4-x^2}\,dx\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
"Shells" is much easier.
We have: \(\displaystyle \L\:2\pi\int^{\;\;\;2}_0 \underbrace{\left(4 - x^2\right)^{^{1/2}}}_{u^{1/2}}\,\underbrace{x\,dx}_{\text{almost }du}\)
Hello, mathhelp!
I was going to comment that the washer method is dangerous
. . near the top of the region.
Then I realized that the graph is drawn incorrectly.
The graph of: \(\displaystyle \,y\:=\:\sqrt{4\,-\,x^2}\,\) is a quarter-circle in Quadrant I.
Code:
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-2 | 2The volume can be found either way:
Washers: \(\displaystyle \L\:V \:=\:\pi\int^{\;\;\;2}_0\left(\sqrt{4-y^2} \,+\,2\right)^2\,dy\)
. . Shells: \(\displaystyle \L\:V \:=\:2\pi\int^{\;\;\;2}_0 x\sqrt{4-x^2}\,dx\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
"Shells" is much easier.
We have: \(\displaystyle \L\:2\pi\int^{\;\;\;2}_0 \underbrace{\left(4 - x^2\right)^{^{1/2}}}_{u^{1/2}}\,\underbrace{x\,dx}_{\text{almost }du}\)
Hey Soroban. If I may, are you sure about your shells?. It is revolved about x=-2. I believe the washers and shells you posted result in two different solutions.
I used :\(\displaystyle \L\\{\pi}\int_{0}^{2}\left[(-2-\sqrt{4-y^{2}})^{2}-(-2)^{2}\right]dy\) for the washers. This gives me the same answer as my shells method.
I believe the first one you have would be the answer if we revolved about y=-2?.
Or is it me?.
Probably is.
I used :\(\displaystyle \L\\{\pi}\int_{0}^{2}\left[(-2-\sqrt{4-y^{2}})^{2}-(-2)^{2}\right]dy\) for the washers. This gives me the same answer as my shells method.
I believe the first one you have would be the answer if we revolved about y=-2?.
Or is it me?.
Probably is.
