5
$\begingroup$

Hello maths community!

One day I was solving a geometry problem and I thought I had found a way of solving it. When I was solving the problem, I kind of invented a new way of finding an area of a shape that is related to circle but just recently I realized that I was wrong. The method I invented was invalid but I just don't understand why wouldn't it work... I won't describe the problem here but I am going to show an example of the method I invented.

Warning. The following methods are invalid!

To find the area of a circle with my method is to first split the circle in half and first find the area of that and then multiply it by 2 in the end to make our life easier.

A half circle is made out of bunch of vertical "lines" and I thought one could use integrals to add these "lines" up to get the area. To do this we need to find a function to get the length of one "line" and then integrate the function. If we draw a right triangle inside the half circle, the task will become easy.

constructing a half circle

triangle inside the half circle

Now, to get $l$, we use sine: $l=r\cdot \sin(φ)$

Then we take the integral with respect to $φ$. I guess this part is the one causing problems...

$$\int_{0}^\pi l \, dφ=\int_0^\pi r\cdot \sin(φ) \, dφ=r\int_0^\pi \sin(φ) \, dφ = r (-\cos(\pi) +\cos(0))=2r$$

Well we know already that the area of a half circle is definitely $2r$ and if we multiplied it by $2$ we would get $4r$ and $4r \neq \pi r^2$

With my logic this would seem very valid and I would just like to know why this doesn't work. Thanks!

$\endgroup$
  • $\begingroup$ What do you mean by integrating wrt $\phi $ Is this a summation process? If so then adding lengths isn't going to have the units for area. $\endgroup$ – Karl Feb 20 '16 at 22:29
  • $\begingroup$ Yeah, I thought adding lines would have worked... $\endgroup$ – Miksu Feb 22 '16 at 5:05
2
$\begingroup$

You need $\displaystyle \int \ell\,dx$, not $\displaystyle\int\ell\,d\varphi$. You have $x=r\cos\varphi$, so $dx = -r\sin\varphi\,d\varphi$.

$\endgroup$
3
$\begingroup$

Integrating means to sum rectangle areas, not lines. The height of a rectangle is $y=r\sin\phi$, but its base is given by the difference in $x$ when the angle varies between $\phi$ and $\phi+\Delta\phi$, that is $\Delta x = -r(\cos(\phi+\Delta\phi)-\cos\phi)$. As you want to take the limit for $\Delta\phi\to0$ you can approximate $\Delta x$ to first order in $\Delta\phi$ as $\Delta x = r\sin\phi\Delta\phi$, so that your sum becomes $$ \lim_{\Delta\phi\to0}\sum r\sin\phi\cdot r\sin\phi\,\Delta\phi= \int_0^\pi r^2\sin^2\phi\, d\phi={\pi\over2}r^2. $$

$\endgroup$
2
$\begingroup$

The problem is the same if you try to measure the area of a $30$-$60$-$90$-degree right triangle by integrating along the hypotenuse, using the length of a line segment parallel to the longer leg. Let $\theta$ be the distance traveled along the hypotenuse, $y$ the length of the line segment from the point at distance $\theta$ to the shorter leg, parallel to the longer leg. Then $y = \frac{\sqrt3}{2}\theta$. If the length of the hypotenuse is $2$, then naïvely we might try to integrate $$ \int_0^2 \frac{\sqrt3}{2}\theta \,d\theta = \left.\frac{\sqrt3}{4}\theta^2 \right|_{\theta=0}^{\theta=2} = \sqrt 3. $$ But the actual area of that triangle is $\frac{\sqrt3}{2}$. What went wrong?

Integration works because we are actually adding up the areas between the line segments. The area between any two line segments can be approximated by a rectangle with length equal to the length of one of the segments and width equal to the distance between them, measured perpendicular to the segments. What is wrong with the naïve approach is that it assumed the distance along the hypotenuse was the distance between segments, but the hypotenuse is not perpendicular to the segments.

There are at least two ways to fix this. One way is to choose a different variable over which to integrate, such as distance along a leg of the triangle. The other way is to account for the ratio between the rate at which we put vertical distance between the segments and the rate at which our integration variable increases. Integrating along the hypotenuse as we did, each unit of travel along the hypotenuse gets us only $\frac12$ unit closer to the opposite leg, and puts only $\frac12$ unit distance between the line segments. So we need an extra factor of $\frac12$ in the integral:

$$ \int_0^2 \frac{\sqrt3}{2}\theta \times \frac12\,d\theta = \left.\frac{\sqrt3}{8}\theta^2 \right|_{\theta=0}^{\theta=2} = \frac{\sqrt3}{2}. $$

For your integration of the circle, the direction of travel along the circumference is constantly changing, and so is the ratio of perpendicular distance gained to the change in $\phi$. This ratio is near zero when $\phi$ is near zero, increases to $1$ when $\phi = \frac\pi2$ (a right angle), and decreases back to zero as $\phi$ approaches $\pi$; in fact, it is possible to show that the ratio is just $\sin\phi$. So the correct integral is $$ \int_{0}^\pi l d\phi=\int_{0}^\pi r \sin(\phi) \times \sin(\phi)\,d\phi, $$ which has already been shown (in another answer) to give the correct area.

$\endgroup$
2
$\begingroup$

One way to find the are, you can use double integral, note that I used polar coordinates $$\int_{\circ}1\,dx\,dy=\int_{0}^{2\pi}\int_0^Rr\,dr\,d\theta=\int_0^R2\pi r \, dr =\pi R^2$$


Another methode:

$$\frac 1 2 \oint_{\gamma} x \, dy- y\,dx \text{ where } \gamma(\theta)=(r \cos \theta, r \sin \theta),\quad 0\leq\theta < 2\pi$$

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

$\endgroup$

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