Question

The value of the line integral \( \int_{C} \left( 2xy^2 \, dx + 2x^2 y \, dy - \frac{1}{3} z \, dz \right) \) along a path joining the origin \( (0,0,0) \) and the point \( (1,1,1) \) is ...........

• A. $\frac{1}{2}$
• B. $\frac{7}{6}$
• C. $\frac{3}{2}$
• D. $\frac{5}{6}$

Hint:

For a line integral over a straight path, parametrize the path and substitute the differentials ($dx$, $dy$, $dz$) accordingly. Integrate each term and substitute the limits to obtain the final result.

Answer 1

The Correct Option is D

We are given the line integral: \[ I = \int_C \left( 2xy^2 \, dx + 2x^2 y \, dy - \frac{1}{3} z \, dz \right) \] The path joining the origin $(0,0,0)$ to the point $(1,1,1)$ is a straight line, so we parametrize the path using: \[ x = t, y = t, z = t \text{for} t \in [0,1] \] Thus, $dx = dt$, $dy = dt$, and $dz = dt$. Now, substitute these into the line integral: \[ I = \int_0^1 \left( 2t \cdot t^2 \, dt + 2t^2 \cdot t \, dt - \frac{1}{3} \cdot t \, dt \right) \] Simplifying the terms inside the integral: \[ I = \int_0^1 \left( 2t^3 \, dt + 2t^3 \, dt - \frac{1}{3} t \, dt \right) \] \[ I = \int_0^1 \left( 4t^3 - \frac{1}{3} t \right) dt \] Now, integrate each term: \[ I = \left[ \frac{4t^4}{4} - \frac{1}{3} \cdot \frac{t^2}{2} \right]_0^1 \] \[ I = \left[ t^4 - \frac{t^2}{6} \right]_0^1 \] Substituting the limits: \[ I = \left[ 1^4 - \frac{1^2}{6} \right] - \left[ 0^4 - \frac{0^2}{6} \right] \] \[ I = 1 - \frac{1}{6} = \frac{5}{6} \] Thus, the value of the line integral is $\frac{5}{6}$.