Question

Let \(C\) be the positively oriented boundary of the domain bounded by the curves \(y=2x^{2}\) and \(y^{2}=4x\). Then the value of the line integral \[ \oint_{C} (2y^{2}+2xy+4y)\,dx + (x^{2}+4xy+8x)\,dy \] is equal to:

• A. \(\dfrac{8}{3}\)
• B. \(\dfrac{2}{3}\)
• C. \(\dfrac{4}{3}\)
• D. \(\dfrac{1}{3}\)

Hint:

Whenever the integrand is of the form \(P\,dx+Q\,dy\) over a closed curve, check Green’s theorem. If \(\partial Q/\partial x-\partial P/\partial y\) is constant, the integral equals that constant times the enclosed area—often much simpler than a direct path integral.

Answer 1

The Correct Option is A

Step 1: Identify \(P(x,y)\) and \(Q(x,y)\) for Green’s theorem.
Set \[ P(x,y)=2y^{2}+2xy+4y,\qquad Q(x,y)=x^{2}+4xy+8x. \] By Green’s theorem, \[ \oint_{C}P\,dx+Q\,dy=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA, \] where \(D\) is the region enclosed by \(C\).

Step 2: Compute the curl term.
\[ \frac{\partial Q}{\partial x}=2x+4y+8,\qquad \frac{\partial P}{\partial y}=4y+2x+4. \] Hence \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =(2x+4y+8)-(4y+2x+4)=4. \] So the integral reduces to \[ \oint_C P\,dx+Q\,dy=\iint_D 4\,dA=4\cdot \text{Area}(D). \]

Step 3: Find the area of the region \(D\).
The curves intersect where \(y=2x^{2}\) and \(y^{2}=4x\). Substitute \(y=2x^{2}\) into \(y^{2}=4x\): \[ (2x^{2})^{2}=4x\ \Rightarrow\ 4x^{4}=4x\ \Rightarrow\ x(x^{3}-1)=0 \Rightarrow x=0 \text{ or } x=1. \] Thus intersection points are \((0,0)\) and \((1,2)\). For \(0\le x\le 1\), the upper curve is \(y=2\sqrt{x}\) (from \(y^{2}=4x\)) and the lower curve is \(y=2x^{2}\). Therefore, \[ \text{Area}(D)=\int_{0}^{1}\big(2\sqrt{x}-2x^{2}\big)\,dx =2\!\left[\frac{2}{3}-\frac{1}{3}\right] =\frac{2}{3}. \]

Step 4: Evaluate the line integral.
\[ \oint_C P\,dx+Q\,dy=4\cdot \text{Area}(D)=4\cdot \frac{2}{3}=\frac{8}{3}. \] Final Answer: \(\boxed{\dfrac{8}{3}}\)