Question

If 𝐴 = 3𝑥 2 + 6𝑦 𝑖 − 14𝑦𝑧𝑗 + 20𝑥𝑧 2𝑘, ��׬ ��evaluate 𝐴 ∙ 𝑑𝑟 from (0,0,0) to (1,1,1) along the following path C: 𝑎. 𝑥 = 𝑡, 𝑦 = 𝑡 2 , 𝑧 = 𝑡 3 b. the straight lines from (0,0,0) to (1,0,0) and then to (1,1,0) and the to (1, 1, 1) c. the straight line joining (0,0,0) to (1,1,1).

Answer 1

(a) Give, $x=t,y=t^2,z=t^3$

The points corresponds to t=1

$dx=dt,dy=2tdt,dz=3t^2dt \\[9pt] \int \vec{A}.\vec{dr}=\int(3x^2+6y)dx-14yzdy+20xz^2dz \\[9pt] =\int_0^1(3t^2+6t^2)dt-14t^5(2tdt)+20t^7(3t^2dt) \\[9pt] =5$

(b)

$\int_{(0,0,0,)}^{(1,0,0)} \vec{A}.\vec{dr}=\int_0^1(3x^2+6y)dx-\int_0^014yzdy+\int_0^020xz^2dz \\[9pt]=x^3+6xy|_0^1=1$

$\int_{(1,1,0,)}^{(1,1,1)} \vec{A}.\vec{dr}=\int_1^1(3x^2+6y)dx-\int_1^114yzdy+\int_0^120xz^2dz \\[9pt]=20\dfrac{xz^3}{3}|_0^1=0$

(c)

$\int_{(0,0,0,)}^{(1,1,1)} \vec{A}.\vec{dr}=\int_0^1(3x^2+6y)dx-\int_0^114yzdy+\int_0^120xz^2dz \\[9pt]=x^3+6xy|_0^1-7y^2z|_0^1+\dfrac{20xz^3}{3}|_0^1\\[9pt]=1+6-7+\dfrac{20}{3}=\dfrac{20}{3}$