Answer to Question #104208 in Electricity and Magnetism for Akshay

Question #104208
Determine the work done by the force
F = (xy+3z)I + (2y² - x²)j + (z - 2y)k in taking a particular from X =0 to X=1 along a curve defined by the equations:
X² = 2y; 2x³ = 3z
1
Expert's answer
2020-03-02T10:14:37-0500

Find the work along


y=x22,  z=23x3y=\frac{x^2}{2},\space\space z=\frac{2}{3}x^3

from x=0 to x=1:


A=LXdx+Ydy+Zdz==01(xy+3z)dx+(2y2x2)dy+(z2y)dz.A=\int^LXdx+Ydy+Zdz=\\ =\int^1_0(xy+3z)dx+(2y^2 - x^2)dy+(z - 2y)dz.

From the first pair of expressions, we have


dy=xdx,  dz=2x2dx.dy=xdx,\space\space dz=2x^2dx.

Substitute this, as well as the first pair of equations:


A=01(xx22+32x33)dx+ +[2(x22)2x2]xdx+ +(2x332x22)2x2dx= =01[5x32+x52x3+4x532x4]dx= =01[11x562x4+3x32]dx= =11x6362x55+3x4801= =(1116362155+3148)0=101360.A=\int^1_0(x\cdot\frac{x^2}{2}+3\cdot\frac{2x^3}{3})dx+\\ \space\\+\bigg[2\bigg(\frac{x^2}{2}\bigg)^2 - x^2\bigg]xdx+\\ \space\\ +\bigg(\frac{2x^3}{3} - 2\cdot\frac{x^2}{2}\bigg)2x^2dx=\\ \space\\ =\int^1_0\bigg[\frac{5x^3}{2}+\frac{x^5}{2}-x^3+\frac{4x^5}{3}-2x^4\bigg]dx=\\ \space\\ =\int^1_0\bigg[\frac{11x^5}{6}-2x^4+\frac{3x^3}{2}\bigg]dx=\\ \space\\ =\frac{11x^6}{36}-\frac{2x^5}{5}+\frac{3x^4}{8}\bigg|^1_0=\\ \space\\=\bigg(\frac{11\cdot1^6}{36}-\frac{2\cdot1^5}{5}+\frac{3\cdot1^4}{8}\bigg)-0=\frac{101}{360}.


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