1. **Problem statement:** Find the work done by the force field \(\vec{F} = (x^3 - 3xy^2) \vec{i} + (3x^2 y - y^3) \vec{j}\) moving an object from \(A(1,0)\) to \(B(0,1)\) along the unit circle \(x^2 + y^2 = 1\). 2. **Work done by a force field along a curve:** The work done is given by the line integral $$ W = \int_C \vec{F} \cdot d\vec{r} = \int_C P\,dx + Q\,dy $$ where \(P = x^3 - 3xy^2\) and \(Q = 3x^2 y - y^3\). 3. **Parameterize the curve:** The path is the unit circle from \(A(1,0)\) to \(B(0,1)\) along the first quadrant. Parameterize by \(x = \cos t, y = \sin t\) with \(t\) from \(0\) to \(\frac{\pi}{2}\). 4. **Compute differentials:** $$ dx = -\sin t \, dt, \quad dy = \cos t \, dt $$ 5. **Substitute into \(P\) and \(Q\):** \[ P = (\cos t)^3 - 3 \cos t (\sin t)^2 = \cos^3 t - 3 \cos t \sin^2 t \] \[ Q = 3 (\cos t)^2 \sin t - (\sin t)^3 = 3 \cos^2 t \sin t - \sin^3 t \] 6. **Calculate the integrand:** $$ P \, dx + Q \, dy = (\cos^3 t - 3 \cos t \sin^2 t)(-\sin t) dt + (3 \cos^2 t \sin t - \sin^3 t)(\cos t) dt $$ 7. **Simplify the expression:** \[ = -\cos^3 t \sin t + 3 \cos t \sin^3 t + 3 \cos^3 t \sin t - \cos t \sin^3 t \] \[ = (-\cos^3 t \sin t + 3 \cos^3 t \sin t) + (3 \cos t \sin^3 t - \cos t \sin^3 t) \] \[ = 2 \cos^3 t \sin t + 2 \cos t \sin^3 t \] \[ = 2 \cos t \sin t (\cos^2 t + \sin^2 t) = 2 \cos t \sin t \] 8. **Integrate over \(t\) from 0 to \(\frac{\pi}{2}\):** $$ W = \int_0^{\pi/2} 2 \cos t \sin t \, dt $$ Use identity \(\sin 2t = 2 \sin t \cos t\), so $$ W = \int_0^{\pi/2} \sin 2t \, dt = \left[-\frac{\cos 2t}{2}\right]_0^{\pi/2} = \frac{1}{2} (\cos 0 - \cos \pi) = \frac{1}{2} (1 - (-1)) = 1 $$ **Final answer for part (a):** \(W = 1\). --- **Slug:** "work force" **Subject:** "vector calculus" **Desmos:** {"latex":"y=\sqrt{1-x^2}","features":{"intercepts":true,"extrema":true}} **q_count:** 3 (Note: Only part (a) is solved as per instructions.)