UPSC Maths 2025 Paper 1 Q7b — Solution

15 marks · Section B

Question

Verify Green’s theorem in the plane for $\displaystyle\oint_C [(xy + y^2)\,dx + x^2\,dy]$ , where $C$ is the boundary of the region bounded by the curves $y = x$ and $y = x^2$ .

Technique

Apply Green’s theorem $\displaystyle\oint_C (P\,dx + Q\,dy) = \iint_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$ by computing both sides independently and checking they agree.

Solution

Here $P = xy + y^2$ , $Q = x^2$ . The curves $y=x$ and $y=x^2$ intersect where $x = x^2$ , i.e. at $(0,0)$ and $(1,1)$ . On $[0,1]$ , $x \ge x^2$ , so the region is $R = \{(x,y): 0\le x\le 1,\ x^2\le y\le x\}.$ The positively-oriented (counterclockwise) boundary $C$ consists of:

$C_1$ : the parabola $y=x^2$ from $(0,0)$ to $(1,1)$ (lower boundary, left→right);
$C_2$ : the line $y=x$ from $(1,1)$ back to $(0,0)$ (upper boundary, right→left).

Double integral (RHS)

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - (x + 2y) = x - 2y.$ $\iint_R (x - 2y)\,dA = \int_0^1\!\!\int_{x^2}^{x}(x - 2y)\,dy\,dx.$ Inner integral: $\int_{x^2}^{x}(x - 2y)\,dy = \bigl[xy - y^2\bigr]_{y=x^2}^{y=x} = (x\cdot x - x^2) - (x\cdot x^2 - x^4) = 0 - (x^3 - x^4) = x^4 - x^3.$ Then $\int_0^1 (x^4 - x^3)\,dx = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}.$

Line integral (LHS)

Along $C_1$ : $y = x^2$ , $dy = 2x\,dx$ , $x:0\to1$ . $\int_{C_1} = \int_0^1\Bigl[(x\cdot x^2 + x^4) + x^2(2x)\Bigr]dx = \int_0^1 (x^3 + x^4 + 2x^3)\,dx = \int_0^1 (3x^3 + x^4)\,dx = \frac{3}{4} + \frac{1}{5} = \frac{19}{20}.$

Along $C_2$ : $y = x$ , $dy = dx$ , $x:1\to0$ . $\int_{C_2} = \int_1^0\Bigl[(x\cdot x + x^2) + x^2\Bigr]dx = \int_1^0 (x^2 + x^2 + x^2)\,dx = \int_1^0 3x^2\,dx = \bigl[x^3\bigr]_1^0 = -1.$

Sum: $\oint_C = \frac{19}{20} + (-1) = \frac{19 - 20}{20} = -\frac{1}{20}.$

Conclusion

$\oint_C (P\,dx + Q\,dy) = -\frac{1}{20} = \iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$ Green’s theorem is verified.

Answer

$\boxed{\;\text{Both sides equal } -\dfrac{1}{20}\ \Rightarrow\ \text{Green's theorem is verified.}\;}$