Question

number 4

Answer 1

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

Answer 2

ellipsoid ?

Answer 3

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.

Answer 4

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?

Answer 5

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

Answer 6

\[\large\rm \color{orangered}{x=\frac{1}{2}u}\]
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:

Answer 7

oh no D:

Answer 8

x= 2u

y= 3v
z= z

Answer 9

Ok good.
We'll call z something new, even though we're not applying a coefficient to it.

(x, y, z) -> (2u, 3v, w)

Answer 10

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

Answer 11

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.

Answer 12

okie

Answer 13

Since we're making a simple linear substitution, our Jacobian will be something simple like this, ya?

Remember how to do that?

Answer 14

hmm oo yes

Answer 15

thats what u did right

Answer 16

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.

Answer 17

oh yeh

Answer 18

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

Answer 19

ah okay that makes sense ;p so then xD

Answer 20

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

Answer 21

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

Answer 22

And now do spherical coordinate business...

Answer 23

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

Answer 24

is the answer 96pi?

Answer 25

Hmm I think it should be 32pi

Answer 26

ohhh lol how u get 32pi

Answer 27

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

Answer 28

lool

Answer 29

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?

Answer 30

oh wait i think i Sauceed up

Answer 31

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

Answer 32

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

Answer 33

Whu 0_o

Answer 34

Where did that equation come from?

Answer 35

wait thats the one we created right or no

Answer 36

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

Answer 37

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

Answer 38

ohh lol

Answer 39

okie so then

Answer 40

\[\large\rm \color{orangered}{u=r \sin \phi \cos \theta}\]\[\large\rm \color{royalblue}{dV=r^2 \sin \theta~dr~d \phi~d \theta}\]

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

Answer 41

And then integrate some stuff.

Answer 42

okie what are the bounds lol

Answer 43

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

Answer 44

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

Answer 45

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