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!

  • 1
    $\begingroup$ your procedure seems to be right, but 4*12=48 not 36. $\endgroup$ – MrYouMath May 16 '16 at 17:33

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

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

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



  • $\begingroup$ oh my goodness... this is why I could never be a math major. Thank you! That's correct $\endgroup$ – Kommander Kitten May 16 '16 at 17:43

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, $$ \int_C\mathbf F\cdot d\mathbf r=\phi\circ\mathbf r\left(\frac\pi2\right)-\phi\circ\mathbf r\left(0\right) = \frac12(\|(3,6,5)\|^2-\|(1,1,1)\|^2) = \frac{67}2. $$


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