UsukiDoll
  • UsukiDoll
Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x^2+y^2 =4 and the plane z + y =3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
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UsukiDoll
  • UsukiDoll
I got all the way to trying to integrate in terms of y
UsukiDoll
  • UsukiDoll
would a simple substitution work?

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anonymous
  • anonymous
What is the first octant? Everything is positive?
anonymous
  • anonymous
Are you using rotation of an sheet, or multiple integrals?
UsukiDoll
  • UsukiDoll
I think everything is positive...
UsukiDoll
  • UsukiDoll
multiple integrals
UsukiDoll
  • UsukiDoll
dx dy that sort of thing
UsukiDoll
  • UsukiDoll
double integrals over general regions...mmmm love the graphing hate the switching
UsukiDoll
  • UsukiDoll
but this problem doesn't involve reversing the integral
UsukiDoll
  • UsukiDoll
first octant is the portion of the xyz-axis system in which all three variables are positive
UsukiDoll
  • UsukiDoll
but I have a circle with c (0,0) and radius 2
anonymous
  • anonymous
Find out where t he circle and line intersect
anonymous
  • anonymous
Actually the first thing we want to do is draw a picture.
UsukiDoll
  • UsukiDoll
|dw:1378703117499:dw|
UsukiDoll
  • UsukiDoll
|dw:1378703181434:dw|
anonymous
  • anonymous
|dw:1378703113231:dw|
UsukiDoll
  • UsukiDoll
|dw:1378703291366:dw|
UsukiDoll
  • UsukiDoll
I see a lot of intersections.....or it may be just me tired...
anonymous
  • anonymous
I think it is bounded above by that plane.
UsukiDoll
  • UsukiDoll
z = 3-y
UsukiDoll
  • UsukiDoll
3-y dx dy
anonymous
  • anonymous
\[ \int_0^2\int_0^{\sqrt{1-x^2}} \int_0^{3-y}dzdydz \]
UsukiDoll
  • UsukiDoll
WHAT THE! Triple integrals? I didn't even learn that yet.... but I do know that it's integrate in terms of z and then y and x
UsukiDoll
  • UsukiDoll
:OOOOOOOO
UsukiDoll
  • UsukiDoll
so that would be z F(3-y) - f( 0) 3-y dy dx
UsukiDoll
  • UsukiDoll
3y-y^2/2
UsukiDoll
  • UsukiDoll
F(sqroot<1-x^2) -f(0) 3sqroot(1-x^2)- [sqroot(1-x^2)]^2/2
UsukiDoll
  • UsukiDoll
|dw:1378703822239:dw|
UsukiDoll
  • UsukiDoll
and then integrate with respect to x... f(2)-f(0)
UsukiDoll
  • UsukiDoll
why are there two dz?
anonymous
  • anonymous
\[ \int_0^2\int_0^{\sqrt{2-x^2}} \int_0^{3-y}dzdydx \]
anonymous
  • anonymous
\[ \int_0^2\int_0^{\sqrt{2-x^2}} \int_0^{3-y}dzdydx = \int_0^2\int_0^{\sqrt{2-x^2}}3-y\;dydx \]
anonymous
  • anonymous
\[ =\int_0^23(\sqrt{2-x^2})-\frac 12 (\sqrt{2-x^2})^2\;dx \]
anonymous
  • anonymous
This integral is a pain in the retrice can you take it from here or do you need help?
UsukiDoll
  • UsukiDoll
back....is that one of those table of intergrals?
UsukiDoll
  • UsukiDoll
If I use simple subsitutuoin u=2-x^2 and du would be 2x but I don't see a placeholder for the x on the integral
UsukiDoll
  • UsukiDoll
maybe factoring one of the sqroots out may work.... reverse product rule???integration by parts
anonymous
  • anonymous
You want to use \(x=\sqrt{2}\sin\theta\)
anonymous
  • anonymous
You integrate the terms separately.
anonymous
  • anonymous
Oh pellet...
anonymous
  • anonymous
It should be \(\sqrt{4-x^2}\)
UsukiDoll
  • UsukiDoll
I see a trig identity coming along...
anonymous
  • anonymous
\[ \int_0^23(\sqrt{4-x^2})-\frac 12 (\sqrt{4-x^2})^2\;dx \]
anonymous
  • anonymous
You want to do the other term separately, because it is easy.
UsukiDoll
  • UsukiDoll
so do I plug in the sinpheta radical 2 into the x and use trig identities before taking the antiderivative?
anonymous
  • anonymous
However for the square root, you need to do \[ x=2\sin\theta \]
UsukiDoll
  • UsukiDoll
just plug it in there or replace in its entirety
anonymous
  • anonymous
It's a \(u\) substitution, basically.
anonymous
  • anonymous
\[dx=2\cos\theta \;d\theta\]
anonymous
  • anonymous
It's like a reverse u substitution
UsukiDoll
  • UsukiDoll
I might see where this is coming from ... similar to sin^-1x
anonymous
  • anonymous
Yeah, but you don't actually want to do the inverse.
anonymous
  • anonymous
\[ \int\sqrt{4-x^2}dx = \int \sqrt{4-4\sin^2\theta }(2\cos\theta \;d\theta) \]
anonymous
  • anonymous
We'll ignore the limits, because worrying about them can lead to more errors.
anonymous
  • anonymous
You continue to simplify.
UsukiDoll
  • UsukiDoll
:P
UsukiDoll
  • UsukiDoll
sorry I need to eat hungry
UsukiDoll
  • UsukiDoll
but I see the simple subsitution or the trig identity comes into play
UsukiDoll
  • UsukiDoll
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