What am I supposed to do?

Calculus Level 5

Find the value of

$\displaystyle I=\int_{C} x^{2}dy - y^{2}dx$

where $C: \dfrac{x^{2}}{4} + \dfrac{y^{2}}{9} =1 , 2 \ge x \ge 0 , 3 \ge y \ge 0$ , oriented from $(2,0)$ to $(0,3)$ .

The answer is 20.

1 solution

Otto Bretscher
May 12, 2016

We can parameterize the given oriented segment $C$ of an ellipse by $x=2\cos t$ and $y=3\sin t$ for $0\leq t\leq \frac{\pi}{2}$ , so that $dx=-2\sin t \hspace{1mm}dt$ and $dy=3\cos t\hspace{1mm}dt$ . Now $I=\int_{0}^{\frac{\pi}{2}}\left((4\cos^2t)(3\cos t)+(9\sin^2t)(2\sin t)\right)dt=\boxed{20}$ .

Can't we solve it using green theorum?

Righved K - 5 years, 1 month ago

We could, but that would be rather "forced" as we have to close up the path first and deal with a rather messy double integral. It's straightforward to do it directly.

Let's hope that somebody will post a Green's solution.

Otto Bretscher - 5 years, 1 month ago

Thanks a lot..(actually I myself got messed up with that double integral) I thought whether I am leaving something or the double integral might be easy to integrate...sorry ,thought deeply :P

Righved K - 5 years, 1 month ago

@Righved K – The double integral is $\int_{0}^{2}\int_{0}^{\frac{3}{2}\sqrt{4-x^2}}2(x+y)dydx=20$

Line integrals (and, more generally, integrals on manifolds) are defined in terms of parametrizations. Green's theorem (or, more generally, Stokes' Theorem, which is really just the fundamental theorem of calculus) sometimes offers a shortcut.

Otto Bretscher - 5 years, 1 month ago

What am I supposed to do?

The answer is 20.

1 solution

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