Questions: Find the work done by F=(x^2+y)i+(y^2+x)j+ze^2k over the following paths from (5,0,0) to (5,0,5). a. The line segment x=5, y=0,0 ≤ z ≤ 5 b. The helix r(t)=(5 cos t) i+(5 sin t) j+(5t/2π) k, 0 ≤ t ≤ 2π c. The x-axis from (5,0,0) to (0,0,0) followed by the line z=x, y=0 from (0,0,0) to (5,0,5) a. Find a scalar potential function f for F, such that F=∇f. A. x^3+xy+y^3+ze^z-e^z+C B. (1/3)x^3+xy+(1/3)y^3+z-e^z+C C. (1/3)x^3+x^2y^2+(1/3)y^3+e^z+C D. (1/3)x^3+xy+(1/3)y^3+ze^z-e^z+C E. The vector field F is not conservative.

Solution Steps

To find the scalar potential function \( f \) for \( \mathbf{F} \), we need to ensure that \( \mathbf{F} = \nabla f \). Given \( \mathbf{F} = (x^2 + y) \mathbf{i} + (y^2 + x) \mathbf{j} + ze^z \mathbf{k} \), we need to find \( f \) such that: \[ \frac{\partial f}{\partial x} = x^2 + y \] \[ \frac{\partial f}{\partial y} = y^2 + x \] \[ \frac{\partial f}{\partial z} = ze^z \]

Integrate \( \frac{\partial f}{\partial x} = x^2 + y \) with respect to \( x \): \[ f(x, y, z) = \int (x^2 + y) \, dx = \frac{1}{3}x^3 + xy + g(y, z) \]

Next, differentiate \( f \) with respect to \( y \) and set it equal to \( \frac{\partial f}{\partial y} \): \[ \frac{\partial}{\partial y} \left( \frac{1}{3}x^3 + xy + g(y, z) \right) = y^2 + x \] \[ x + \frac{\partial g}{\partial y} = y^2 + x \] \[ \frac{\partial g}{\partial y} = y^2 \] Integrate \( \frac{\partial g}{\partial y} = y^2 \) with respect to \( y \): \[ g(y, z) = \frac{1}{3}y^3 + h(z) \]

Now, differentiate \( f \) with respect to \( z \) and set it equal to \( \frac{\partial f}{\partial z} \): \[ \frac{\partial}{\partial z} \left( \frac{1}{3}x^3 + xy + \frac{1}{3}y^3 + h(z) \right) = ze^z \] \[ \frac{dh}{dz} = ze^z \] Integrate \( \frac{dh}{dz} = ze^z \) with respect to \( z \): \[ h(z) = ze^z - e^z + C \]

Combine all parts to form the scalar potential function \( f \): \[ f(x, y, z) = \frac{1}{3}x^3 + xy + \frac{1}{3}y^3 + ze^z - e^z + C \]

Final Answer