2

So I was thinking about the formula for integrals: $\int f(x) dx$. But since $y = f(x)$, I started to wonder about $y dx$ and using a parametrisation of $x$ and $y$ in the integral formula to solve definite integrals.

For instance, I considered the area of a circle of radius $r$, where $x = r \cos t$ and $ y = r \sin t$. So I got the definite integral (by saying that $dx =-r \sin t \;dt$) $\int -r^2(\sin^2(t) dt$, evaluated from $0$ to $2\pi$) and got the result $-\pi r^2$.

Granted, the actual area is $\pi r^2$ but they're at least equal in magnitude, so I was wondering if what I did is a valid operation or not?

Edit: Ok, I think my question was mis-interpreted through my own fault. The question is not really centred around finding the area of the circle using integration. I know how to do that, and I have a good grasp of concepts such as line integrals and arc length. I'm really just experimenting with concepts. For a given function, the integral is defined as $\int f(x) dx$. My line of thinking was that since $y = f(x)$, we can rewrite it as $\int y dx$.

Then, we can consider the parametrisations for $x$ and $y$
$x = g(t)$, $y = h(t)$. Then we say $\frac{dx}{dt} = g'(t)$, $dx = g'(t) dt$. Then we sub back into the equation $\int y dx$ to get $\int h(t)\cdot g'(t) dt$. I did this with the case of a circle with radius $r$ above.

My question is, is it valid to do what I just did, replace x and y with parametric equations? I don't mean this as a way to solve a given integral, but a different way of solving for things such as the area of a circle.

Cataline
  • 2,018
  • 2
    All that is unclear ! – idm Jun 06 '15 at 18:49
  • Unfortunately, I don't know how to write symbols and formulae properly on this website, including how to add the S integral sign. But the underlying question is, can I replace x's and y's with parameteric equations for them in an integral? I base it on the idea that when evaluating a normal integral, it is of the form y dx (I don't know how to add integral sign). – Cataline Jun 06 '15 at 18:56
  • a look at http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference can be helpful ! – Emilio Novati Jun 06 '15 at 19:39

2 Answers2

2

The formula $\int f(x) \mathrm{d} x$ for the area of a region assumes certain properties of $f(x)$. What you have tried to do here is to re-parametrize the equation $x^2+y^2=r^2$. The trouble with doing that for the entire circle is that you end up ignoring the role played by the two different functions $y=\sqrt{r^2-x^2}$ and $y=-\sqrt{r^2-x^2}$ representing the top and bottom half of the circle.

In general -- re-parametrizing is a useful approach to handle areas of the regions where standard antiderivatives cannot be used. This is especially true if the re-parametrization yields an easier integral, such as in your example.

However, you need to be careful and cannot automatically switch into $t$ or any other parameter without examining the objects involved in the process.

I would strongly recommend looking at topics such as line integrals, arc lengths and parametric descriptions of standard functions. These are usually found towards the ends of standard Calculus 2 courses (or at the beginning of Calculus 3 courses) in US universities.

OnceUponACrinoid
  • 772
2

Hint:

If you take the equation $y=\sqrt{r^2-x^2}$ that represents a semicircle, you find the area as: $$ \int_{-r}^r\sqrt{r^2-x^2} dx $$ Your parametric equation is equivalent to the substitution $ x=r\cos t$ but, when you perform such substitution you have to be careful to the integral limits that become: $$ \int_{-r}^r\sqrt{r^2-x^2} dx=\int_{\pi}^0 -r^2\sin^2 t dt $$ and itegrating you have the correct result $\pi r^2/2$

So, using parametric equations can be a good idea, but you have to use these substution with attemption to the limits of integration.

Emilio Novati
  • 62,675