# find the area enclosed by the given ellipse

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.

• You should use double integrals – MysteryGuy May 3 '17 at 9:03
• – MysteryGuy May 3 '17 at 9:11
• 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. – Yves Daoust May 3 '17 at 9:52

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.

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 $$\int_0^{2\pi} \frac{r(\theta)^2}{2} d\theta = \frac{a^2b^2}{2}\int_0^{2\pi} \frac{1}{a^2\cos^2(\theta)+b^2\sin^2(\theta)} d\theta = \pi a b$$

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

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