Question

A Green's Theorem problem, posted below shortly.

Answer 1

\[\text{Calculate} \int\limits_{C}^{}x^2y^3dx+2x^3y^2dy \ \text{where C is the boundary of the region}\]\[x^2+y^2 \leq 1, \ \ \ x,y \geq 0 \ \text{oriented counterclockwise.}\]

Answer 2

I'm terrible at line integrals at the moment, so I'm going to start with part B and then move onto part A; part B is using Green's Theorem, part A is directly.

Answer 3

\[\int\limits_{C}^{}x^2y^3dx+2x^3y^2dy=\int\limits_{}^{}\int\limits_{}^{}N_{x}-M_{y} \ dA\]

Answer 4

you may use `iint` for double integral

Answer 5

Thank you! Heh.

Answer 6

How do you phrase the two separate integrals bounds for \iint?

Answer 7

you use \int_ \int_ when you have bounds
use `iint` only when you dont have bounds

Answer 8

\[\iint N_{x}-M_{y}=\iint 6x^2y^2-3x^2y^2 \ dA = \iint 3x^2y^2 \ dA\]

Answer 9

i wish the integrand is 0 !

Answer 10

\[\iint 3x^2y^2 \ dA = \int\limits_{\text{stuff goes here}}^{\text{pls help}}\int\limits_{\text{some garbage lol}}^{\text{what is this I don't even}}\]

Answer 11

Okay, so, uh, bounds of integration, should I immediately convert to polar, or what?

Answer 12

T-minus 7 hours, 14 minutes. Godspeed, future me.

Answer 13

yes immediately convert to polar, jus tkeep in mind the region is only in first quadrant

Answer 14

Whether you want it or not, here's an intuitive look into Green's Theorem, which is just Stokes theorem in 2D instead of 3D, hopefully it will allow you to "see" what's going on while solving the problem. The idea behind Green's theorem is simple: Any time you put two line integrals next to each other and are opposite they cancel each other out.

The other thing you have to understand is that the curl of a vector gives this sort of output: So to get everything to rotate in the same direction, we either need to change dy/dx or dx/dy to go with it. I'm using the terms x and y a little ambiguously I know, so perhaps replace it with dP/dy and dQ/dx.

The point of this interpretation is that evaluating the curl at a point is the same thing as taking the line integral around that point.

So what we do now is because of the first thing I described, that line integrals next to each other cancel, then if we integrate the curl over the whole shape all the inside pieces cancel each other out leaving just the outside piece:

I hope that helps, I'm not sure maybe it isn't what you want or care about haha.