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.
$\endgroup$ 24 Answers
$\begingroup$Hint: scale $x$ and $y$ suitably and you get a circle. What does that do to the area?
$\endgroup$ 1 $\begingroup$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 $A= \pi a b$.
$\endgroup$ 2 $\begingroup$$$ \text{ Area = }\int y dx = \int_0^{2\pi} y(\theta) dx(\theta) = \int_0^{2\pi} -ab \sin^2 \theta d\theta = \pi ab $$
$\endgroup$ $\begingroup$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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