Compute the line integral of v = 6 ˆx + yz^2 ˆy + (3y + z) ˆz along the triangular path shown in Fig. 1.49. Check your answer using Stokes’ theorem. [Answer: 8/3]

Question

Compute the line integral of

[latex]v = 6 \hat{x} + yz^2 \hat{y} + (3y + z)\hat{z} [/latex]

along the triangular path shown in Fig. 1.49. Check your answer using Stokes’ theorem. [Answer: 8/3]

Accepted Answer

(1) [latex]x = z = 0; dx = dz = 0; y : 0 → 1. v · dl = (yz^2) dy = 0; \int{}v · dl = 0[/latex] .

(2) [latex]x = 0; z = 2−2y; dz = −2 dy; y : 1 → 0. v·dl = (yz^2) dy+(3y+z) dz = y(2−2y)^2 dy−(3y+2−2y)2 dy[/latex];

[latex]\int{}v · dl = 2\int\limits_{1}^{0}{}(2y^3 − 4y^2 + y − 2) dy = 2 \left(\frac{y^4}{2}-\frac{4y^3}{3} +\frac{y^2}{2} -2y ight)\mid ^{0}_{1} =\frac{14}{3} [/latex] .

(3) x = y = 0; dx = dy = 0; z : 2 → 0. v · dl = (3y + z) dz = z dz;

[latex]\int{}v · dl =\int\limits_{2}^{0}{}z dz =\frac{z^2}{2} \mid ^{0}_{2}= −2 [/latex] .

Total: [latex]\oint{}v · dl = 0+\frac{14}{3} −2 =\frac{8}{3} [/latex] .

Meanwhile, Stokes’ thereom says [latex]\oint{}v · dl =\int{}(∇ imes v) [/latex] . da. Here [latex]da = dy dz \hat{x}[/latex] , so all we need is

[latex](∇ imes v)_x =\frac{\partial}{\partial y}(3y + z) −\frac{\partial}{\partial z}(yz^2) = 3 − 2yz[/latex] . Therefore

[latex]\int{}(∇ imes v) · da =\iint{}(3 − 2yz) dy dz =\int_{0}^{1}{} \left[\int_{0}^{2-2y}{} (3 − 2yz) dz ight]dy [/latex]

=[latex]\int_{0}^{1}{}\left[3(2 − 2y) − 2y\frac{1}{2}(2 − 2y)^2 ight]dy =\int_{0}^{1}{}(−4y^3 + 8y^2 − 10y + 6) dy [/latex]

=[latex]\left(−y^4 +\frac{8}{3}y^3 − 5y^2 + 6y ight)\mid ^{1}_{0}= −1 +\frac{8}{3}− 5 + 6 =\frac{8}{3} [/latex] .

Question 1.56: Compute the line integral of v = 6 ˆx + yz^2 ˆy + (3y + z) ˆ...

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