← 2025 Paper 1

UPSC 2025 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

Line integrals · Vector Analysis · asked 8× in 14 yrs · Read the full method →

Question

Verify Green’s theorem in the plane for ∮C[(xy+y2) dx+x2 dy]\displaystyle\oint_C [(xy + y^2)\,dx + x^2\,dy], where CC is the boundary of the region bounded by the curves y=xy = x and y=x2y = x^2.

Technique

Apply Green’s theorem ∮C(P dx+Q dy)=∬R(∂Q∂x−∂P∂y)dA\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+y2P = xy + y^2, Q=x2Q = x^2. The curves y=xy=x and y=x2y=x^2 intersect where x=x2x = x^2, i.e. at (0,0)(0,0) and (1,1)(1,1). On [0,1][0,1], x≥x2x \ge x^2, so the region is

R={(x,y):0≤x≤1, x2≤y≤x}.R = \{(x,y): 0\le x\le 1,\ x^2\le y\le x\}.

The positively-oriented (counterclockwise) boundary CC consists of:

Double integral (RHS)

∂Q∂x−∂P∂y=2x−(x+2y)=x−2y.\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - (x + 2y) = x - 2y. ∬R(x−2y) dA=∫01 ⁣ ⁣∫x2x(x−2y) dy dx.\iint_R (x - 2y)\,dA = \int_0^1\!\!\int_{x^2}^{x}(x - 2y)\,dy\,dx.

Inner integral:

∫x2x(x−2y) dy=[xy−y2]y=x2y=x=(x⋅x−x2)−(x⋅x2−x4)=0−(x3−x4)=x4−x3.\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

∫01(x4−x3) dx=15−14=−120.\int_0^1 (x^4 - x^3)\,dx = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}.

Line integral (LHS)

Along C1C_1: y=x2y = x^2, dy=2x dxdy = 2x\,dx, x:0→1x:0\to1.

∫C1=∫01[(x⋅x2+x4)+x2(2x)]dx=∫01(x3+x4+2x3) dx=∫01(3x3+x4) dx=34+15=1920.\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 C2C_2: y=xy = x, dy=dxdy = dx, x:1→0x:1\to0.

∫C2=∫10[(x⋅x+x2)+x2]dx=∫10(x2+x2+x2) dx=∫103x2 dx=[x3]10=−1.\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:

∮C=1920+(−1)=19−2020=−120.\oint_C = \frac{19}{20} + (-1) = \frac{19 - 20}{20} = -\frac{1}{20}.

Conclusion

∮C(P dx+Q dy)=−120=∬R(∂Q∂x−∂P∂y)dA.\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

  Both sides equal −120 ⇒ Green’s theorem is verified.  \boxed{\;\text{Both sides equal } -\dfrac{1}{20}\ \Rightarrow\ \text{Green's theorem is verified.}\;}
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