# Green's theorem, relation between two integrals

Homework Statement .
Calculate by a line integral the following double integral:
##\iint\limits_D (y^{2}e^{xy}-x^{2}e^{xy})dxdy##, D being the unit disk.

The attempt at a solution.

Well, if we consider C to be the curve that encloses the region D (C is the unit circle), then C is a simple, closed and smooth curve. Now, if we take ##F=(P(x,y),Q(x,y))##
with ##P(x,y)=-ye^{xy}## and ##Q(x,y)=-xe^{xy}##, then P and Q are ##C^1## functions. We are under the hypothesis of Green's theorem, so we know that the integral

##\iint\limits_D (Q_x-P_y)dxdy## equals the integral ##\int\limits_C Fds## where C is the unit circle. If I parametrize this curve in the standard way, I get ##ψ(t)=(cos(t),sin(t))## and ##ψ'(t)=(-sin(t),cos(t))## so the integral of the right member is
##\int_0^{2π} (-sin(t)e^{cos(t)sin(t)},-cos(t)e^{cos(t)sin(t)})(-sin(t),cos(t))
dt##, which is equal to ##\int_0^{2π} sin^{2}(t)e^{cos(t)sin(t)}-cos^{2}(t)e^{cos(t)sin(t)} dt##
I am totally stuck with this integral, I could express it by ##\int_0^{2π}e{cos(t)sin(t)}-2cos^{2}(t)e^{cos(t)sin(t)} dt## but I don't know how to solve it. Am I applying Green's theorem correctly or did I make a mistake in some previous step?

SteamKing
Staff Emeritus
Homework Helper
You've got a basic problem with your analysis and selection of the functions P and Q.

When you take your partial derivatives of P and Q and calculate (Qx - Py), you must obtain the original integrand of the double integral, namely (y^2EXP(xy) - x^2EXP(xy)). Have you checked this?

• 1 person
Dick
Homework Helper
Homework Statement .
Calculate by a line integral the following double integral:
##\iint\limits_D (y^{2}e^{xy}-x^{2}e^{xy})dxdy##, D being the unit disk.

The attempt at a solution.

Well, if we consider C to be the curve that encloses the region D (C is the unit circle), then C is a simple, closed and smooth curve. Now, if we take ##F=(P(x,y),Q(x,y))##
with ##P(x,y)=-ye^{xy}## and ##Q(x,y)=-xe^{xy}##, then P and Q are ##C^1## functions. We are under the hypothesis of Green's theorem, so we know that the integral

##\iint\limits_D (Q_x-P_y)dxdy## equals the integral ##\int\limits_C Fds## where C is the unit circle. If I parametrize this curve in the standard way, I get ##ψ(t)=(cos(t),sin(t))## and ##ψ'(t)=(-sin(t),cos(t))## so the integral of the right member is
##\int_0^{2π} (-sin(t)e^{cos(t)sin(t)},-cos(t)e^{cos(t)sin(t)})(-sin(t),cos(t))
dt##, which is equal to ##\int_0^{2π} sin^{2}(t)e^{cos(t)sin(t)}-cos^{2}(t)e^{cos(t)sin(t)} dt##
I am totally stuck with this integral, I could express it by ##\int_0^{2π}e{cos(t)sin(t)}-2cos^{2}(t)e^{cos(t)sin(t)} dt## but I don't know how to solve it. Am I applying Green's theorem correctly or did I make a mistake in some previous step?

Use double angle formulas. Doesn't it sort of resemble cos(2t)e^(sin(2t))?

• 1 person
You've got a basic problem with your analysis and selection of the functions P and Q.

When you take your partial derivatives of P and Q and calculate (Qx - Py), you must obtain the original integrand of the double integral, namely (y^2EXP(xy) - x^2EXP(xy)). Have you checked this?

I see, P and Q were not the correct ones. I have to set ##P=xe^{xy}## and ##Q=ye^{xy}##. I spent like 15 minutes trying to find where my mistake was and I totally missed it.

Use double angle formulas. Doesn't it sort of resemble cos(2t)e^(sin(2t))?

The integral is 0, my problem was I hadn't chosen P and Q appropriately.

SteamKing
Staff Emeritus