Answer to Question #108052 in Field Theory for ZEESHAN FAZAL

Question

Determine the work done by the force F(xy+3z)i^+(2y^2 -x^2)j+(z-2y)k in taking a particle from  x = 0 to  x =1 along a curve defined by the equations: x^2= 2y ; 2x^3= 3z

Answer 1

$$\frac{dy}{dx}=x,\frac{dz}{dx}=2x^2$$

$$W=\int_0^1((xy+3z)+(2y^2 -x^2)x+(z-2y)2x^2)dx$$

$$W=\int_0^1Idx$$

$$(xy+3z)=x\frac{x^2}{2}+2x^3=\frac{5x^3}{2}$$

$$(2y^2 -x^2)x=\left(2\left(\frac{x^2}{2}\right)^2 -x^2\right)x=\left(\frac{x^5}{2} -x^3\right)$$

$$(z-2y)2x^2=\left(\frac{2x^3}{3}-x^2\right)2x^2=\left(\frac{4x^5}{3}-2x^4\right)$$

$$I=\frac{5x^3}{2}+\left(\frac{x^5}{2} -x^3\right)+\left(\frac{4x^5}{3}-2x^4\right)=\left(\frac{11x^5}{6}+\frac{3x^3}{2}-2x^4\right)$$

$$W=\int_0^1\left(\frac{11x^5}{6}+\frac{3x^3}{2}-2x^4\right)dx$$

$$W=\left(\frac{11}{6(6)}+\frac{3}{2(4)}-2\frac{1}{5}\right)=\frac{101}{360}\approx 0.28$$