0
$\begingroup$

Find the area enclosed by the given ellipse:

$$ (x,y)=(a \cos t, b \sin t) \: , \quad 0\leq t < 2\pi $$

I have tried to google this as well as look in my notes but I don't know where to start. please point me in the right direction.

$\endgroup$
  • $\begingroup$ You should use double integrals $\endgroup$ – MysteryGuy May 3 '17 at 9:03
  • 2
    $\begingroup$ Does it help : math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap9/section2/… $\endgroup$ – MysteryGuy May 3 '17 at 9:11
  • $\begingroup$ You can stretch space by horizontal and vertical factors $1/a$ and $1/b$ so that the ellipse transforms into the unit circle, known to have area $\pi$. As the stretching transforms scale the area in the same proportion, the initial area is $\pi ab$. Note that the trick will not work for the perimeter, as the stretching transforms do not scale the arc lengths. $\endgroup$ – Yves Daoust May 3 '17 at 9:52
2
$\begingroup$

Use good old $$\int y\,dx.$$

By symmetry, you can integrate on the half-ellipse, with $t$ from $\pi$ to $0$ (i.e. $x$ from $-a$ to $a$). We have

$$\frac A2=\int_\pi^0 a\sin t\,(-b\sin t)\,dt=ab\int_0^\pi\sin^2t\,dt.$$

As$$\sin^2t=\frac{1-\cos2t}2$$ the value of the integral is $\dfrac\pi2$, giving

$$A=\pi ab.$$


Note that integration from $2\pi$ to $0$ directly yields the correct answer, as it performs a forward pass with $y$, and a backward pass, which is equivalent to a forward pass with $-y$, because $y\,dx=(-y)(-dx)$. Hence you are integrating $y_+(x)-y_-(x)$ where $y_+,y_-$ are the upper and lower arcs. This is a general method that works for closed curves. The reversal of the range, $2\pi$ to $0$, ensures clockwise rotation.

$\endgroup$
1
$\begingroup$

Using Green's theorem, area is given by $$\iint_A 1\, dA = \frac{1}{2}\oint_ C x \,dy-y\, dx = \frac{1}{2}\int_0^{2\pi} ab\cos^2(t)+ab\sin^2(t)\, dt = \pi ab$$

Alternatively, if we define a coordinate system such that $x=r\cos(\theta), y=b\frac{r}{a}\sin(\theta)$, then the Jacobian turns out to be $J=br/a$, and so the area is given by $$\int_0^{2\pi}\int_0^a \frac{b}{a}r \, dr\, d\theta = \pi ab$$

Finally, if you want to use single variable calculus, note that the polar equation for an ellipse is $$r(\theta) = \frac{ab}{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}}$$ and so area is given by $$\int_0^{2\pi} \frac{r(\theta)^2}{2} d\theta = \frac{a^2b^2}{2}\int_0^{2\pi} \frac{1}{a^2\cos^2(\theta)+b^2\sin^2(\theta)} d\theta = \pi a b$$

$\endgroup$
0
$\begingroup$

You have the ellipse equation $x^2/a^2 + y^2/b^2 = 1$. The area of this ellipse is well known: $A = \pi a b$.

$\endgroup$
-1
$\begingroup$

area is $\pi ab$ also you can derive the area using definite integration

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.