Line Integrals of Vector Fields โ€” Question 1

Section 18-4 ยท Calculus pages 3401โ€“3402

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Question 1

Let ๐‘ญ(x,y)=โŸจy,x2โŸฉ,๐’“(t)=โŸจt,t2โŸฉ,0โ‰คtโ‰ค1.\mathbf F(x,y)=\langle y,x^2\rangle,\qquad \mathbf r(t)=\langle t,t^2\rangle,\quad 0\le t\le 1. Evaluate the line integral โˆซC๐‘ญโ‹…d๐’“\displaystyle\int_C\mathbf F\cdot d\mathbf r.

Tasks

  1. Evaluate ๐‘ญ\mathbf F along the curve and compute ๐’“โ€ฒ(t)\mathbf r'(t).

  2. Form the dot product and evaluate the integral.

  3. State how the sign of the integrand agrees with the geometry.

Question 1 โ€“ Solution

Strategy. Use โˆซC๐‘ญโ‹…d๐’“=โˆซab๐‘ญ(๐’“(t))โ‹…๐’“โ€ฒ(t)dt\int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt.

Step 1: Restrict the field to the curve Since x=tx=t and y=t2y=t^2, ๐‘ญ(๐’“(t))=โŸจt2,t2โŸฉ,๐’“โ€ฒ(t)=โŸจ1,2tโŸฉ.\mathbf F(\mathbf r(t))=\langle t^2,t^2\rangle,\qquad \mathbf r'(t)=\langle 1,2t\rangle.

Diagram shown in the original SVG worksheet.

Step 2: Form the scalar integrand ๐‘ญ(๐’“(t))โ‹…๐’“โ€ฒ(t)=t2+2t3.\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=t^2+2t^3. Therefore โˆซC๐‘ญโ‹…d๐’“=โˆซ01(t2+2t3)dt=[t33+t42]01=56.\int_C\mathbf F\cdot d\mathbf r =\int_0^1(t^2+2t^3)\,dt =\left[\frac{t^3}{3}+\frac{t^4}{2}\right]_0^1 =\boxed{\frac 56}.

Verification For 0โ‰คtโ‰ค10\le t\le 1, both components of the field and both components of the velocity are nonnegative. Their dot product must be nonnegative, consistent with the positive answer.