Evaluate the Line Integral by Applying Green's Theorem: Four Closed Curves

Answer

Each curve below is positively oriented and bounds a single region, so Green's theorem converts the line integral into a double integral over that region:

$\oint_{C} P\,dx+Q\,dy=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$

146. Region between $y=x^{3}$ and $y=x$: $-\frac{1}{60}$

With $P=2xy$ and $Q=x+y$, the partial derivatives are $\frac{\partial Q}{\partial x}=1$ and $\frac{\partial P}{\partial y}=2x$, so the double integrand is $1-2x$.

The path goes from $(0,0)$ to $(1,1)$ along $y=x^{3}$ and returns along $y=x$. On $0\le x\le 1$ the curve $y=x^{3}$ lies below $y=x$, so the region is $0\le x\le 1$, $x^{3}\le y\le x$:

$\oint_{C}2xy\,dx+(x+y)dy=\int_{0}^{1}\int_{x^{3}}^{x}(1-2x)\,dy\,dx=\int_{0}^{1}(1-2x)\left(x-x^{3}\right)dx.$

$\int_{0}^{1}\left(x-x^{3}-2x^{2}+2x^{4}\right)dx=\frac{1}{2}-\frac{1}{4}-\frac{2}{3}+\frac{2}{5}=-\frac{1}{60}.$

The answer is negative even though the traversal is counterclockwise: the integrand $1-2x$ is negative wherever $x>\frac{1}{2}$, and on this thin region the negative part wins.

147. Region between $y=0$ and $y=4-x^{2}$: $\frac{32}{3}$

The same field gives the same integrand $1-2x$. The parabola meets $y=0$ at $x=\pm 2$, so $D$ is $-2\le x\le 2$, $0\le y\le 4-x^{2}$ and

$\iint_{D}(1-2x)dA=\iint_{D}1\,dA-2\iint_{D}x\,dA.$

The first integral is the area of the region, $\int_{-2}^{2}(4-x^{2})dx=\frac{32}{3}$. The second is zero, because $D$ is symmetric about the $y$-axis while $x$ is odd. The line integral is therefore $\frac{32}{3}$.

148. Ellipse $x=4+2\cos\theta$, $y=4\sin\theta$: $0$

Here $P=2\arctan\!\left(\frac{y}{x}\right)$ and $Q=\ln(x^{2}+y^{2})$. The ellipse is centred at $(4,0)$ with semi-axes $2$ and $4$, so every point of the ellipse and of its interior has $x\ge 2>0$: neither component has a singularity inside the region, and the required partial derivatives are continuous there.

$\frac{\partial P}{\partial y}=2\cdot\frac{1}{1+(y/x)^{2}}\cdot\frac{1}{x}=\frac{2x}{x^{2}+y^{2}},\qquad \frac{\partial Q}{\partial x}=\frac{2x}{x^{2}+y^{2}}.$

The two partial derivatives agree everywhere on the region, so the double integral is $0$ and the line integral is $0$. The integrand is the differential of an exact form on the right half-plane, so the result is the same for any positively oriented simple closed curve that stays in $x>0$.

149. Region between $y=x$ and $y=\sqrt{x}$: $\frac{1}{12}$

With $P=\sin x\cos y$ and $Q=xy+\cos x\sin y$,

$\frac{\partial Q}{\partial x}=y-\sin x\sin y,\qquad \frac{\partial P}{\partial y}=-\sin x\sin y,\qquad \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=y.$

On $0\le x\le 1$ the curve $y=x$ lies below $y=\sqrt{x}$, so $D$ is $0\le x\le 1$, $x\le y\le\sqrt{x}$:

$\iint_{D}y\,dA=\int_{0}^{1}\int_{x}^{\sqrt{x}}y\,dy\,dx=\int_{0}^{1}\frac{x-x^{2}}{2}dx=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{12}.$

The trigonometric parts cancel exactly in $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$, which is what leaves the short double integral in $y$ alone.

What to check before applying the theorem

• Positive orientation. Traversing the same curve clockwise reverses the sign of every answer above.
• A simple closed curve. The theorem as used here needs a curve that does not cross itself and that bounds exactly one region.
• The field must be smooth on the whole region, not merely on the curve. When a component is undefined inside the enclosed region, the double integral has to be split and treated as an improper integral — or the theorem cannot be used at all.