Calculating work done by Vector Field on particle moving on curve

Question

I've been working on this problem for an hour now, and I'm still not getting the right answer. Here is the problem:

If C is the curve given by $$r(t)=(1+2\sin(t))i+(1+5\sin^2(t))j+(1+4\sin^3(t))k $$$$ 0≤t≤π/2$$ and $F$ is the radial vector field $$F(x,y,z)=xi+yj+zk$$ compute the work done by F on a particle moving along C.

So work done is given by: $$\int\limits_C {{\bf{F}}\cdot d{\bf{r}}}$$

Our parametrization is given in the problem. Taking the derivative and plugging it in the line integral, we have:

$$\int\limits_C {(1+2\sin(t), 1 + 5\sin^2(t), 1 + 4\sin^3(t)) \cdot (2\cos(t), 10\sin(t)\cos(t), 12\sin^2(t)\cos(t)) dt}$$

Evaluating... $$\int_0^{\pi/2} {2\cos(t) + 14\sin(t)\cos(t) + 50\sin^3(t)\cos(t) + 12\sin^2(t)\cos(t) + 36\sin^5(t)\cos(t)} \, dt$$

Which gives me $31.5$. But this isn't the right answer. Am I approaching the problem correctly? Is my set up of the integral correct? Thank you in advance!

Answer 1

Your calculations are correct up to one before the last integral:

$$\int_0^{\pi/2}\left(2\cos t+2\sin2t+5\sin2t+50\sin^3t\cos t+ 12\sin^2t\cos t+48\sin^5t\cos t\right)dt=$$

$$=\left.\left(2\sin t-\frac72\cos2t+\frac{25}2\sin^4t+4\sin^3t+8\sin^6t\right)\right|_0^{\pi/2}=$$

$$=2(1-0)-\frac72(-1-1)+\frac{25}2(1-0)+4(1-0)+8(1-0)$$

$$2+7+\frac{25}2+4+8=\frac{67}2$$

Answer 2

When you’re given nasty-looking problems like this, there’s often an easier way to solve them than by grinding through them directly. For line integrals, it’s generally a good idea to see if the vector field is conservative before plunging in. If it is, it might be easier to find a scalar potential $\phi(\mathbf r)$ such that $\nabla\phi=\mathbf F$ and then evaluate $\phi$ at the endpoints of the path.

In this case, $\nabla\times\mathbf F=0$, so $\mathbf F$ is indeed conservative. To find $\phi$, you can proceed by successive integration and differentiation, like so: $$ \phi_x=x \implies \phi(x,y,z)=\frac12x^2+g(y,z) \\ \phi_y=g_y=y \implies g(y,z)=\frac12y^2+h(z) \\ \phi_z=h_z=z \implies h(z)=\frac12z^2+C. $$ Putting this all together, we have $$\phi(x,y,z)=\frac12(x^2+y^2+z^2)+C = \frac12\|\mathbf r\|^2+C.$$ We can drop the constant of integration since it’ll cancel out when we subtract the two values of $\phi$ from each other. Finally, $$ \int_C\mathbf F\cdot d\mathbf r=\phi\circ\mathbf r\left(\frac\pi2\right)-\phi\circ\mathbf r\left(0\right) = \frac12(\|(3,6,5)\|^2-\|(1,1,1)\|^2) = \frac{67}2. $$