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I am not sure how to work this one out.

I am suppose to find the area of this parametric equation.

$$y = b\sin\theta, x = a\cos\theta$$

$$0 \leq 0 \leq 2\pi$$

I set up the equation in the memorized formula.

$$\int_0^{2\pi} \sqrt{1 + \left(\frac{b\cos\theta}{-a\sin\theta}\right)^2}d\theta$$

$$\int_0^{2\pi} \sqrt{1 + \frac{b^2\cos^2\theta}{a^2\sin^2\theta}}d\theta$$

$$\int_0^{2\pi} \sqrt{1 + \frac{b^2}{a^2}\cdot \csc^2\theta}d\theta$$

From here I am at a loss of what to do, I tried some trig subsitution but I do not think that will work and it only seems to complicate the problem.

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The formula that you used is for the length of the boundary curve, not for the area. If you want to work out the area, maybe you can compute the area that's added as theta changes by a small amount dtheta. The extra area is approximately a triangle. If you work out the area of the triangle and add these up, you should be in business. –  anthonyquas Jun 25 '12 at 20:56
Please cancel the (differential-equations) tag for this question as this question is not the business of ODE. –  doraemonpaul Jun 26 '12 at 2:34

4 Answers 4

Note that $\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=\cos^2\theta + \sin^2\theta=1.$ We then recognize that this describes an ellipse, which has area $\pi a b$.

EDIT: How to find the area of the ellipse. There are many many proofs of this, but the easiest one you might find in a single-variable calculus course is as follows.

We will find the area of the ellipse in the first quadrant and quadruple it (we assume $x,y\geq0$ for what follows). Solve for $y$ in terms of $x$. $$ 1 = \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 $$ $$ \iff \left(\frac{y}{b}\right)^2= 1-\left(\frac{x}{a}\right)^2 $$ $$ \iff \frac{y}{b} = \sqrt{1-\left(\frac{x}{a}\right)^2} $$ $$ y=b\sqrt{1-\left(\frac{x}{a}\right)^2}. $$ Then we have $$ \frac{A}{4} = \int_0^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx. $$ Substitute $u = \frac{x}{a}$ so $dx=a\ du$ and the limits go from $u=0$ to $u=1$, $$ \frac{A}{4} = ab\int_0^1\sqrt{1-u^2}du. $$ At this point you could interpret the integral on the right as the the area of a quarter of a circle of radius 1, $\pi/4$. If not, we continue by making the trigonometric substitution $u = \sin t$, so $\sqrt{1-u^2}=\cos t$ and $du=\cos t\ dt$ where $t$ goes from $0$ to $\pi/2$, $$ \frac{A}{4} = ab\int_0^{\pi/2} \cos^2t dt. $$ Now we use the formula $\cos^2t = \frac{1}{2}(1+\cos(2t))$, $$ \frac{A}{4} = \frac{ab}{2}\int_0^{\pi/2}(1+\cos(2t))dt $$ $$ =\frac{ab}{2}\left(t+\frac{1}{2}\sin(2t) \right)_0^{\pi/2} $$ $$ =\frac{ab}{2}\left(\left(\frac{\pi}{2}+0\right) - \left(0 + 0\right) \right) $$ $$ \frac{\pi a b}{4}. $$ Rearranging then gives us $A= \pi a b$.

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And how do you prove the ellipse has that area? –  Pedro Tamaroff Jun 26 '12 at 23:47
Edited to include single-variable calculus proof of area of ellipse. –  nullUser Jun 27 '12 at 1:10
(+1) ${}{}{}{}{}$ –  Pedro Tamaroff Jun 27 '12 at 1:10

You used the arclength formula, when you should have used another one.

You know that the area of a function on an interval $(a,b)$ is given by

$$\int_a^b f(x) dx=\int_a^b y\; \; dx$$

Since $y=f(x)$. But since you have that

$$y= a\sin \theta$$ $$x = b \cos\theta$$ you can use that wisely. First note that $ \theta$ ranges from $0$ to $2 \pi$, since the sine and cosine are periodic (which means then you'd just be repeating the graph over itself). So you need

$$\int_0^{2 \pi} a\sin \theta\; \; dx$$

But we still need to replace $dx$, which turns out to be

$$dx = -b \sin\theta d\theta$$ so we get

$$-\int_0^{2 \pi} ab\sin ^2\theta\; \; d\theta$$

In general, you can find the area of $y=y(t)$, $x=x(t)$ by using $$\int_a^b y(t) x'(t) dt$$ as above and choosing $a$ and $b$ appropiately.

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$$ \text{ Area = }\int y dx = \int_0^{2\pi} y(\theta) dx(\theta) = \int_0^{2\pi} -ab \sin^2 \theta d\theta = \pi ab $$

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Hint: scale $x$ and $y$ suitably and you get a circle. What does that do to the area?

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