← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q8c — Step-by-Step Solution

13 marks · Section B

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

Question

Let F⃗=xy2i^+(y+x)j^\vec F=xy^2\hat i+(y+x)\hat j. Integrate (∇×F⃗)⋅k^(\nabla\times\vec F)\cdot\hat k over the region in the first quadrant bounded by the curves y=x2y=x^2 and y=xy=x using Green’s theorem.

Technique

(∇×F⃗)⋅k^=Qx−Py=1−2xy(\nabla\times\vec F)\cdot\hat k=Q_x-P_y=1-2xy; integrate over RR with x2≤y≤xx^2\le y\le x, 0≤x≤10\le x\le1.

Solution

Step 1 — The integrand (∇×F⃗)⋅k^(\nabla\times\vec F)\cdot\hat k

With F⃗=(P,Q)=(xy2, y+x)\vec F=(P,Q)=(xy^2,\ y+x) (a planar field),

(∇×F⃗)⋅k^=∂Q∂x−∂P∂y=∂∂x(y+x)−∂∂y(xy2)=1−2xy.(\nabla\times\vec F)\cdot\hat k=\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =\frac{\partial}{\partial x}(y+x)-\frac{\partial}{\partial y}(xy^2)=1-2xy.

By Green’s theorem this double integral equals the boundary line integral ∮∂R(P dx+Q dy)\oint_{\partial R}(P\,dx+Q\,dy); we evaluate the double integral directly (both are computed below).

Step 2 — Describe the region

The curves y=x2y=x^2 and y=xy=x meet where x2=xx^2=x, i.e. x=0x=0 and x=1x=1. On [0,1][0,1] the line lies above the parabola (x≥x2x\ge x^2), so

R={(x,y):0≤x≤1, x2≤y≤x}.R=\{(x,y):0\le x\le1,\ x^2\le y\le x\}.

Step 3 — Iterated integral

∬R(1−2xy) dA=∫01 ⁣ ⁣∫x2x(1−2xy) dy dx.\iint_R(1-2xy)\,dA=\int_0^1\!\!\int_{x^2}^{x}(1-2xy)\,dy\,dx.

Inner integral:

∫x2x(1−2xy) dy=[y−xy2]y=x2y=x=(x−x⋅x2)−(x2−x⋅x4)=(x−x3)−(x2−x5)=x−x2−x3+x5.\int_{x^2}^{x}(1-2xy)\,dy=\big[y-xy^2\big]_{y=x^2}^{y=x} =(x-x\cdot x^2)-(x^2-x\cdot x^4)=(x-x^3)-(x^2-x^5)=x-x^2-x^3+x^5.

Outer integral:

∫01(x−x2−x3+x5) dx=12−13−14+16.\int_0^1\big(x-x^2-x^3+x^5\big)\,dx=\frac12-\frac13-\frac14+\frac16.

Common denominator 1212: 6−4−3+212=112.\dfrac{6-4-3+2}{12}=\dfrac{1}{12}.

Answer

  ∬R(∇×F⃗)⋅k^ dA=112.  \boxed{\;\iint_R(\nabla\times\vec F)\cdot\hat k\,dA=\frac{1}{12}.\;}
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