# 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?

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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
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.