xXMarcelieXx:

number 4

1 year ago
xXMarcelieXx:
1 year ago

Zepdrix:

What shape is region G? Can you tell by looking at the equation?

1 year ago
xXMarcelieXx:

ellipsoid ?

1 year ago
Zepdrix:

Yes. And we don't want to integrate over an ellipsoid. No bueno. We would prefer to integrate over a sphere. So we'll apply the necessary substitutions to make that happen.

1 year ago
Zepdrix:

We want to change x into this new variable u. How do we get rid of that denominator under the x? What coefficient should we apply to u?

1 year ago
xXMarcelieXx:

im thinking of x= 1/2 u y=1/3 v z= z

1 year ago
Zepdrix:

\[\large\rm \color{orangered}{x=\frac{1}{2}u}\] Hmm let's see what happens to x if we do that. \[\large\rm \frac{\color{orangered}{(x)}^2}{4}+\frac{y^2}{9}+z^2 \le 1\] \[\large\rm \frac{\color{orangered}{\left(\frac{1}{2}u\right)}^2}{4}+\frac{y^2}{9}+z^2 \le 1\] \[\large\rm \frac{u^2}{4\cdot4}+\frac{y^2}{9}+z^2 \le 1\] Hmm that doesn't seemed to have worked. That gave us an extra 1/4 D:

1 year ago
xXMarcelieXx:

oh no D:

1 year ago
xXMarcelieXx:

x= 2u y= 3v z= z

1 year ago
Zepdrix:

Ok good. We'll call z something new, even though we're not applying a coefficient to it. (x, y, z) -> (2u, 3v, w)

1 year ago
Zepdrix:

So now we have this new region S:\[\large\rm u^2+v^2+w^2\le1\]

1 year ago
Zepdrix:

In order to integrate in terms of u,v,w, we'll need to replace the x^2 with something in terms of u, and we'll need to calculate our Jacobian.

1 year ago
xXMarcelieXx:

okie

1 year ago
Zepdrix:

|dw:1511170494028:dw|Since we're making a simple linear substitution, our Jacobian will be something simple like this, ya? Remember how to do that?

1 year ago
xXMarcelieXx:

hmm oo yes

1 year ago
xXMarcelieXx:

thats what u did right |dw:1511170747390:dw|

1 year ago
Zepdrix:

Yes, I sort of skipped ahead and did some of it in my head. Notice with our substitution, x only has u, y only has v, z only has w. x doesn't have any v or w. So the x_v and x_w are 0. We'll only end up with x_u, y_v, z_w. Everything else zeros.

1 year ago
xXMarcelieXx:

oh yeh

1 year ago
Zepdrix:

Whenever you make a direct linear substitution like we did, the Jacobian will just be the product of the coefficients. (x, y, z) -> (2u, 3v, 1w) => J = 2 * 3 * 1 = 6

1 year ago
xXMarcelieXx:

ah okay that makes sense ;p so then xD

1 year ago
Zepdrix:

Ok so we'll substitute the x, and jam our Jacobian in there as well,\[\large\rm \iiint_G x^2 dV_{x,y,z}\quad=\quad \iiint_S (2u)^2~6~dV_{u,v,w}\]

1 year ago
Zepdrix:

\[\large\rm =24\iiint u^2dV\]

1 year ago
Zepdrix:

And now do spherical coordinate business...

1 year ago
Zepdrix:

\(\large\rm u=r \sin\phi\cos\theta\) \(\large\rm dV=r^2\sin\theta~dr~d\phi~d\theta\)

1 year ago
xXMarcelieXx:

is the answer 96pi?

1 year ago
Zepdrix:

Hmm I think it should be 32pi

1 year ago
xXMarcelieXx:

ohhh lol how u get 32pi

1 year ago
Zepdrix:

Oh we're integrating u^2 lol I made boo boo ;p lemme try again hehe

1 year ago
xXMarcelieXx:

lool

1 year ago
Zepdrix:

Mmm I don't remember spherical all that well.. What do you have for your boundaries? 0 <= theta <= 2pi -pi/2 <= phi <= pi/2 0 <= r <= 1 Do those look right?

1 year ago
xXMarcelieXx:

oh wait i think i Sauceed up

1 year ago
Zepdrix:

Sauceed up? x'D x'D x'D

1 year ago
xXMarcelieXx:

x'D lmao i think those boundaries look right so you would look at the new eqaution 2u+3v+w = 1 ?

1 year ago
Zepdrix:

Whu 0_o

1 year ago
Zepdrix:

Where did that equation come from?

1 year ago
xXMarcelieXx:

wait thats the one we created right or no

1 year ago
Zepdrix:

Naw lady. We had an ellipsoid. And the new equation we created was a sphere.

1 year ago
Zepdrix:

\[\large\rm \frac{x^2}{4}+\frac{y^2}{9}+z^2\le1\qquad\qquad\to\qquad\qquad u^2+v^2+w^2\le1\]

1 year ago
xXMarcelieXx:

ohh lol

1 year ago
xXMarcelieXx:

okie so then

1 year ago
Zepdrix:

\[\large\rm \color{orangered}{u=r \sin \phi \cos \theta}\]\[\large\rm \color{royalblue}{dV=r^2 \sin \theta~dr~d \phi~d \theta}\] We make our conversion to spherical coordinates,\[\large\rm 24\iiint (\color{orangered}{u})^2\color{royalblue}{dV}\quad=\quad 24\iiint (\color{orangered}{r \sin \phi \cos \theta})^2\color{royalblue}{r^2 \sin \theta~dr~d \phi~d \theta}\]

1 year ago
Zepdrix:

And then integrate some stuff.

1 year ago
xXMarcelieXx:

okie what are the bounds lol

1 year ago
Zepdrix:

I posted them earlier. I don't really remember the bounds for a full circle. I think it's like this: 0 <= theta <= 2pi -pi/2 <= phi <= pi/2 0 <= r <= 1

1 year ago
sweetburger:

phi is bound from 0 to pi for a sphere theta covers the angles that go past pi

1 year ago
sweetburger:

also note that the angle phi is with respect to the z-axis

1 year ago