Question

If \(\vec F=3xy\hat i-y^2\hat j\), then the value of \(\int_C \vec F\cdot d\vec r\), where \(C\) is the curve \(y=2x^2\) in the \(xy\)-plane from \((0,0)\) to \((1,2)\), is

• A. \(-\dfrac{7}{6}\)
• B. \(\dfrac{7}{6}\)
• C. \(0\)
• D. \(\dfrac{6}{7}\)

Hint:

For line integrals in the plane, write \(\vec F\cdot d\vec r=Pdx+Qdy\), then use the curve equation to convert everything into one variable.

Answer 1

The Correct Option is A

Concept:
For a vector field \[ \vec F=P\hat i+Q\hat j \] the line integral is \[ \int_C \vec F\cdot d\vec r=\int_C P\,dx+Q\,dy \] Here, \[ P=3xy,\qquad Q=-y^2 \]

Step 1: Use the curve equation. The curve is \[ y=2x^2 \] Differentiate: \[ dy=4x\,dx \] The curve goes from \((0,0)\) to \((1,2)\), so \[ x:0\to 1 \]

Step 2: Substitute in the line integral. \[ \int_C \vec F\cdot d\vec r = \int_C 3xy\,dx-y^2\,dy \] Using \[ y=2x^2 \] we get \[ 3xy=3x(2x^2)=6x^3 \] and \[ y^2=(2x^2)^2=4x^4 \] Also, \[ dy=4x\,dx \] So, \[ -y^2dy=-(4x^4)(4x\,dx) \] \[ =-16x^5\,dx \] Therefore, \[ \int_C \vec F\cdot d\vec r = \int_0^1 (6x^3-16x^5)\,dx \]

Step 3: Integrate. \[ \int_0^1 6x^3\,dx=\left[\frac{6x^4}{4}\right]_0^1=\frac{3}{2} \] \[ \int_0^1 16x^5\,dx=\left[\frac{16x^6}{6}\right]_0^1=\frac{8}{3} \] So, \[ \int_0^1(6x^3-16x^5)\,dx = \frac{3}{2}-\frac{8}{3} \] \[ =\frac{9-16}{6} \] \[ =-\frac{7}{6} \]

Step 4: Final answer. \[ \boxed{-\frac{7}{6}} \]