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!
Best Answer
Your calculations are correct up to one before the last integral:
$$\int_0^{\pi/2}\left(2\cos t+2\sin2t+5\sin2t+50\sin^3t\cos t+ 12\sin^2t\cos t+48\sin^5t\cos t\right)dt=$$
$$=\left.\left(2\sin t-\frac72\cos2t+\frac{25}2\sin^4t+4\sin^3t+8\sin^6t\right)\right|_0^{\pi/2}=$$
$$=2(1-0)-\frac72(-1-1)+\frac{25}2(1-0)+4(1-0)+8(1-0)$$
$$2+7+\frac{25}2+4+8=\frac{67}2$$