Question

If \( C \) is the boundary of the region enclosed by the curves \( y = \sqrt{x} \) and \( y = x^2 \), then the value of the line integral
\[ \int_C \left( (3x^2 - 8y^2) \, dx + (4y - 6xy) \, dy \right) = ............ \]

• A. $\dfrac{7}{2}$
• B. $\dfrac{3}{2}$
• C. $\dfrac{3}{10}$
• D. $\dfrac{7}{10}$

Hint:

Green's Theorem is useful for converting a line integral into a double integral over the region enclosed by the curve. Make sure to compute the partial derivatives correctly when applying the theorem.

Answer 1

The Correct Option is B

To solve the line integral, we use Green's Theorem, which relates a line integral over a closed curve $C$ to a double integral over the region $R$ enclosed by $C$: \[ \int_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where $P(x, y) = 3x^2 - 8y^2$ and $Q(x, y) = 4y - 6xy$.
1. First, compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4y - 6xy) = -6y \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3x^2 - 8y^2) = -16y \] 2. Now, apply Green's Theorem: \[ \int_C \left( (3x^2 - 8y^2) \, dx + (4y - 6xy) \, dy \right) = \iint_R \left( -6y - (-16y) \right) \, dA \] \[ = \iint_R 10y \, dA \] 3. The region $R$ is enclosed by the curves $y = \sqrt{x}$ and $y = x^2$. We set up the limits for the double integral:
- $x$ varies from 0 to 1 (the points where the curves intersect).
- For each $x$, $y$ varies from $x^2$ to $\sqrt{x}$.
Thus, the double integral becomes: \[ \int_0^1 \int_{x^2}^{\sqrt{x}} 10y \, dy \, dx \] 4. Solve the inner integral: \[ \int_{x^2}^{\sqrt{x}} 10y \, dy = 10 \left[ \frac{y^2}{2} \right]_{x^2}^{\sqrt{x}} = 10 \left( \frac{x}{2} - \frac{x^4}{2} \right) = 5(x - x^4) \] 5. Now, solve the outer integral: \[ \int_0^1 5(x - x^4) \, dx = 5 \left( \int_0^1 x \, dx - \int_0^1 x^4 \, dx \right) \] \[ = 5 \left( \frac{1}{2} - \frac{1}{5} \right) = 5 \times \frac{3}{10} = \frac{15}{10} = \frac{3}{2} \] Thus, the value of the line integral is $\dfrac{3}{2}$.