GATE 2023 EC Q.37: solution and answer

Official answer: BSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Write the integrand as $\vec{F}\cdot d\vec{r}$ with $\vec{F}=(z^2,3y^2,2xz)$ and test whether the field is conservative. The curl components are $\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}=0-0=0$, $\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}=2z-2z=0$ and $\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=0-0=0$. So $\nabla \times \vec{F}=\vec{0}$ everywhere in ${\mathbb{R}}^3$: the field is conservative and the integral does not depend on the path.

2. Find the potential. $z^{2}dx+2xzdz=d(x{z}^2)$ and $3y^{2}dy=d(y^3)$, so $\varphi (x,y,z)=x{z}^2+y^3$. Check: $\nabla \varphi =(z^2,3y^2,2xz)=\vec{F}$.

3. By the fundamental theorem for line integrals, ${\int }_P^Q\nabla \varphi \cdot d\vec{r}=\varphi (Q)-\varphi (P)$. Here $\varphi (Q)=\varphi (2,3,1)=2\cdot 1^2+3^3=29$ and $\varphi (P)=\varphi (1,1,2)=1\cdot 2^2+1^3=5$.

4. So the integral is $29-5=24$.

5. Check along the given straight line: $x=1+t,y=1+2t,z=2-t$ for $0\le t\le 1$, so $dx=dt$, $dy=2dt$, $dz=-dt$. Then $${\int }_0^1\left[(2-t{)}^2+6(1+2t{)}^2-2(1+t)(2-t)\right]dt=\frac{7}{3}+26-\frac{13}{3}=24$$ which agrees.

6. The value is $24$, which is option (B).