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I am currently working on understanding this problem and am in need of some assistance. I'll let you know what I've done so far, and then hopefully someone will be able to help me.

So the problem is as follows:

Give a parametrization to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1, a,b > 0 and write down the range of your parameter. Then use this parametrization to find the area this ellipse encloses.

Alright so I believe I've gotten the correct parametrization with the following

x = a cos t

y = b sin t

And the formula for the area is: $\int y dx$

The only thing I'm stuck on is how to find the correct range of parameters and how to evaluate the integral. I believe there is some sort of trick to figuring out the integral because if you try to evaluate it normally the top will cancel the bottom and you'll end up with zero.

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Parametrization you intend is $x=a\cos t$, $\dots$. It doesn't cancel. – André Nicolas Apr 29 '12 at 18:16
yeah your right I'll change it, good catch. Do you have any other tips to complete the problem? – Math_Illiterate Apr 29 '12 at 18:40
For what range of $t$ does your parameterization go once around the ellipse? Write down what you have, that allegedly cancels. – GEdgar Apr 29 '12 at 18:44
@GEdgar I'm not sure what you mean. If you plug in my parametrization the $a^2$ and $b^2$ will cancel I believe, leaving $(sinx)^2 + (cosx)^2$ – Math_Illiterate Apr 29 '12 at 18:52
@user28694: It is clear that we go $0$ to $2\pi$. You end up integrating $-ab\sin^2 t$, so there can't be cancellation, but there is that silly negative sign, mostly because of wrong formula for area. The integration can be done many ways, maybe $\cos 2t=1-2\sin^2 t$ is fastest. – André Nicolas Apr 29 '12 at 18:52
up vote 0 down vote accepted

One way to calculate the area is to use the integral $\int y \,dx$ when $x$ changes from $-a$ to $a$. This corresponds to $t$ changing from $\pi$ to $0$ and results in the area of the upper half. For both halves, multiply by $2$:

\begin{align*} \mathcal{A} &= 2 \int_{\pi}^{0} (b\sin{t}) (-a\sin{t}) \,dt \\ &= 2ab \int_{\pi}^{0} -\sin^2{t} \,dt \\ &= 2ab \int_{\pi}^{0} \left(-\frac{1}{2} + \frac{1}{2}\cos(2t)\right) \,dt \\ &= 2ab \left[-\frac{t}{2} + \frac{1}{4}\sin(2t) \right]_{\pi}^{0} \\ &= \pi ab \end{align*}

You'll get the same result if you integrate $-\int y \,dx$ for $t \in [0, 2\pi]$. See Green's Theorem for justification.

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