Question

Green's theorem...

Answer 1

Use Green’s theorem to evaluate

\[\int\limits_{C}^{}(x^2+y^2)dx+(x^2+2xy)dy\]

where C is the boundary of the region in the first quadrant determined by the graphs of
\[x=0, y=1, y=x^2\]

Not after a straight out answer, but any guidance as to how to do it would be really appreciated :)

Cheers!

Answer 2

Looks like we're studying the same material.....my final exam is tomorrow!

Answer 3

I think you have the formula, right?

Answer 4

Haha hey Nisha me too!!

Answer 5

AnwarA... which formula? :S

Answer 6

i just solved one like this last night.....see if I can find it

Answer 7

So evaluate those partial derivatives and determine the boundaries for x and y and start integrating.

Answer 8

Ok, here:
\[\int\limits_{C}^{}P dx+Qdy=\int\limits_{D}^{}\int\limits_{}^{}({\partial Q \over \partial x}-{\partial P \over \partial y})dA.\]
The only problematic part here is to find the boundaries of the region D. Everything else is straight forward. I'll do it in steps in the next comment.

Answer 9

never mind, I messed my integral up, anwarA is right.

Answer 10

But isn't \[\delta(x^2+y^2)/\delta x = \delta(x^2+2xy)/\delta y = 2x\]
... so it would be zero? :S

Answer 11

ooooooooh k hang on a sec then haha

Answer 12

In our case, we have \(P=x^2+y^2 \implies \frac{\partial P}{\partial y}=2y\), \(Q=x^2+2xy \implies \frac{\partial Q}{\partial x}=2x+2y\), and \(dA=dx dy\). Now, we only have to find the boundaries, which is NOT difficult at all. But it's better to be done using graphs.

Answer 13

ahh greens thoerm , nice and easy

Answer 14

Are both the limits 0 to 1...? Or am I just a bit special...?

Answer 15

draw up the region

Answer 16

you can either have { 0<=x<=1 , x^2<=y<=1}

Answer 17

You can see from the graph that \(0 \le x \le 1\) and \(x^2 \le y \le 1\).

Answer 18

or {0<=y<=1 , 0<=x<=sqrt{y} }

Answer 19

the first description of the region is the best to work with , so use that

Answer 20

Now to sum up everything, our desirable integral (call it I) is going to be:
\[I=\int\limits_{0}^{1}\int\limits_{x^2}^{1} (2x+2y-2y)dy dx=\int\limits_{0}^{1}\int\limits_{x^2}^{1}2x dy dx\]

Answer 21

That integral is very easy and I found it equal to \(\frac{1}{2}\).

Answer 22

I am not sure of you're even reading what I have been writing :D

Answer 23

Sheesh, I was reading it :-) Thank you!

Answer 24

multivariable calculus is pretty fun I reckon

Answer 25

It is, more problems please :)

Answer 26

I think it is too!

Answer 27

oh, but I don't understand this surface area stuff

Answer 28

sketch the solid bounded by x^2 +y^2 =4 , z=1 and z=9-3x-2y . Calculate its volume.

Answer 29

if you want another question

Answer 30

Oh, I made a mistake in the boundaries. Check it again; it should be \( 0\le y \le 1\) and \( 0 \le x \le \sqrt{y}\). Sorry for that :(

Answer 31

no, it doesnt matter

Answer 32

they both describe the same region .

Answer 33

Not really :)

Answer 34

never mind, again.

Answer 35

It's easier to make mistakes here than when you're solving in papers!!

Answer 36

true

Answer 37

z=x^2+y^2 ?

Answer 38

Find the surface area of the surface S. S is the paraboloid x^2 + y ^ 2- z = 0 below the plane z = 99/4

Answer 39

\[x=rcos(\theta )\]\[y= rsin(\theta)\]

z= r^2

Answer 40

so 0<=r^2<=99/5 , 0<= theta <= 2 pi

Answer 41

the first inequality comes frm the range on z , and should be 0<=r^2<= 99/4

Answer 42

so\[0\le r \le \frac{\sqrt{99}}{2}\]

Answer 43

now you need a normal vector

then normal vector is defined as the cross product of \[\frac{df}{d \theta} \times \frac{df}{dr}\]

Answer 44

where \[f = ( x(r, \theta) , y(r, \theta), z( r, \theta ) ) \]

Answer 45

take the cross product of the vectors \[( -r \sin(\theta) , rcos(\theta) , 0 ) \times ( \cos(\theta) , \sin(\theta) , 2r ) \]

Answer 46

find that cross product , then find its length ( which will be a function of r and theta ) , and then integrate that over the rectangular region :\[0 \le r \le \frac{ \sqrt{99}}{2}\]\[0 \le \theta \le 2 \pi \]

Answer 47

you can do the rest.

Answer 48

thanks

Answer 49

Hahaha I just went out for some dinner (not sure where you guys are in the world, but for me its now 9:23pm) and couldn't have asked for a better thread to come back to! Thanks guys!