AMU 2026 — Mathematics PYQ

1. Identify the Core Concept: Green's Theorem

The line integral is taken over a simple, closed, counter-clockwise curve $C$ which forms the boundary of a region $R$. Therefore, we can apply Green's Theorem to convert the line integral into a simpler double integral over the region $R$:

$${\oint }_C(Mdx+Ndy)={\iint }_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy$$

From the given problem statement:

• $M=y$

• $N=2x$

2. Compute the Partial Derivatives

Now, let's find the partial derivatives with respect to their corresponding variables:

$$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}(2x)=2$$

$$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}(y)=1$$

Substitute these partial derivatives back into Green's Theorem formula:

$$\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=2-1=1$$

Thus, the integral simplifies to:

$${\oint }_{C}ydx+2xdy={\iint }_{R}1\cdot dxdy$$

Note: ${\iint }_{R}1\cdot dxdy$ simply represents the total geometric Area of Region $R$.

3. Setting Up Limits of Integration

The region $R$ is defined explicitly by the inequalities:

• Outer variable limits: $-1\le y\le 1$

• Inner variable limits: $0\le x\le 1-y^2$

Let's set up the iterated double integral:

$$\text{Area}={\int }_{y=-1}^1{\int }_{x=0}^{1-y^2}dxdy$$

4. Evaluate the Integral Step-by-Step

• Step 4a: Integrate with respect to $x$

  $${\int }_0^{1-y^2}dx=[x{]}_0^{1-y^2}=1-y^2$$

• Step 4b: Integrate with respect to $y$

  Substitute the result back into the outer integral:

  $$\text{Area}={\int }_{-1}^1(1-y^2)dy$$

Since $(1-y^2)$ is an even function (its value stays the same when replacing $y$ with $-y$), we can simplify the integration limits from $[-1,1]$ to $[0,1]$ and multiply by $2$:

$$\text{Area}=2{\int }_0^1(1-y^2)dy$$

Now, find the antiderivative:

$$\text{Area}=2{\left[y-\frac{y^3}{3}\right]}_0^1$$

Evaluate it at the limits $1$ and $0$:

$$\text{Area}=2\left[\left(1-\frac{1^3}{3}\right)-(0-0)\right]$$

$$\text{Area}=2\left[1-\frac{1}{3}\right]=2\left[\frac{2}{3}\right]=\frac{4}{3}$$

Conclusion

The value computed from Green's theorem matches the line integral exactly:

$${\oint }_{C}ydx+2xdy=\frac{4}{3}$$

This directly matches option (d).