Consider the vector field F(x,y)=⟨y,5⟩, and let C be the curve of the ellipse having equation frac{x^{2}}{4}+frac{y^{2}}{9}=1. Compute the line integral ∫CF dr, where r(t) is a parametrization of C going once around counterclockwise.

Consider the vector field F(x,y)=⟨y,5⟩, and let C be the curve of the ellipse having equation frac{x^{2}}{4}+frac{y^{2}}{9}=1. Compute the line integral ∫CF dr, where r(t) is a parametrization of C going once around counterclockwise.

Question
Analysis
asked 2021-02-08
Consider the vector field \(\displaystyle{F}{\left({x},{y}\right)}=⟨{y},{5}⟩,\) and let C be the curve of the ellipse having equation \(\displaystyle{\frac{{{x}^{{{2}}}}}{{{4}}}}+{\frac{{{y}^{{{2}}}}}{{{9}}}}={1}.\) Compute the line integral ∫CF dr, where r(t) is a parametrization of C going once around counterclockwise.

Answers (1)

2021-02-09
The first thing we need to do is come up with a parametrization of the ellipse. In general, you can parameterize the equation
\(\displaystyle{\frac{{{x}^{{{2}}}}}{{{a}^{{{2}}}}}}+{\frac{{{y}^{{{2}}}}}{{{b}^{{{2}}}}}}={1}\)
by x=±acos(ωt+θ0)x=±acos(ωt+θ0), y=±bsin(ωt+θ0)y=±bsin(ωt+θ0) for any angle θθ and angular frequency ωω. We can see this is the case by simply substituting these expressions into the above equation and confirming that it works out. The choice of the sign and phase angles just comes down to which direction you want to go around the ellipse and where you'd like to start on the ellipse. And the choice of ωω just determines how "fast" you want to go around the ellipse. The speed doesn't matter at all for us so let's choose ω=1 so it basically disappears. The sign should be positive if we want to go around counterclockwise. And it really doesn't matter where we start on the ellipse as we just need to make one full trip around so we might as well set θ0=0 as well.
Hence our parametrization becomes x=2cos(t), y=3sin(t). Please confirm a few things about this before moving on: (1) this is a parametrization of the ellipse given by the equation in the question, (2) it goes in the correct direction, and (3) this parametrization is periodic in 2π2π and hence we can take our integral over t=0 to t=2π.
With that all set, we almost ready to substitute in and evaluate this integral. The remaining piece is that we need to differentiate this parametrization vector r(t)=⟨2cos(t),3sin(t)⟩r(t)=⟨. So here we go.
\(\displaystyle{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left[{2}{\cos{{\left({t}\right)}}},{3}{\sin{{\left({t}\right)}}}\right]}={\left[-{2}{\sin{{\left({t}\right)}}},{3}{\cos{{\left({t}\right)}}}\right]}\)
There we go. Now we're set to work on this integral. Don't forget that drdr is basically just shorthand for (dr/dt)dt and hence
\(\displaystyle\int{c}{F}\cdot{d}{r}={\int_{{{t}{0}}}^{{{t}{1}}}}{F}{r}{\left({t}\right)}\times{\frac{{{d}{r}}}{{{t}}}}={\int_{{{0}}}^{{{2}π}}}{\left\lbrace{3}{\sin{{\left({t}\right)}}},{5}\right\rbrace}\times{\left\lbrace-{2}{\sin{{\left({t}\right)}}},{3}{\cos{{\left({t}\right)}}}\right\rbrace}{\left.{d}{t}\right.}={\int_{{{0}}}^{{{2}π}}}-{6}{{\sin}^{{{2}}}{\left({t}\right)}}+{15}{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}=-{6}π\)
Let me know if you need help with the integration. It should be straightforward, just remember to do a power reduction on the sin^2term.
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