# Rectangles in Polar Coordinates

__Circles in rectangular coordinates...Rectangles in polar coordinates?__

This is a bit of a goofy one. Why? Well, I want to figure out something very simple using something much more complicated. There are many examples of this sort of thing - and I just find it to be a great aid in understanding the more complicated idea. So let's figure out the area of a square using double integrals in polar coordinates!

The first time we ever learn the equation of a circle, it's in Cartesian coordinates - aka rectangular coordinates. We have the xy-plane and define the unit circle as the set of all points distance 1 from the origin. Using the distance formula, that means we want all pairs ** (x,y)** with

Squaring both sides and simplifying the expressions gives the equation we all know and love:

Which produces this wonderful circle:

To switch to polar coordinates, we replace ** (x,y) **with

**by remembering that a point on the unit circle has**

*(r,Î¸)***coordinate**

*y***and has**

*sin(Î¸)***coordinate**

*x***. If the circle has radius**

*cos(Î¸)***, then we multiple both trig functions by**

*r***, giving us the change-of-coordinates**

*r**x = rcos(Î¸)*

*y = rsin(Î¸)*

For the other direction, remember that we can calculate the radius via the distance formula, and that ** tan(Î¸)** is defined as

**.**

*y/x*This means that to get the same unit circle in polar coordinates, our equation is much simpler!

*r = 1*

If we set the other coordinate ** Î¸ **to a constant, we get a straight line through the origin with that constant determining its angle. For example,

**gives a line with slope**

*Î¸ = 1***:**

*tan(1)*For contrast, setting our rectangular coordinates to constants, ** x = 1** and

**, gives vertical and horizontal lines.**

*y = 1*We know how to rewrite ** r = 1** and

**in rectangular coordinates, so how do we rewrite**

*Î¸ = 1***and**

*x = 1***in polar coordinates? Well if**

*y = 1***, then**

*x = 1***, so**

*rcosÎ¸ = 1***Similarly, setting**

*r = 1/cosÎ¸ = secÎ¸.***gives**

*y = 1***, so**

*rsinÎ¸ = 1*

*r = cscÎ¸.*We've learned how to draw straight lines in polar coordinates. If we want to draw ** y = 1**, then we use

**. If we only wanted to draw the line segment between**

*r = cscÎ¸***and**

*x = -1***, then we're only interested in the angles between**

*x = 1***and**

*Î¸ = Ï€/4*

*Î¸ = 3Ï€/4.*To draw a rectangle, we just combine these functions and restrictions for each side!

__Bring in the integrals!__

Although it's nice to have the description, but to figure out the area of the square, we really only need one of the functions. If we want a square with side-length ** S**, then we can take the function

**and restrict**

*r = S(cscÎ¸)***to the interval**

*Î¸***. The region this sweeps out is:**

*[Ï€/4, Ï€/2]*A triangle! Thus, the area of the square of side-length ** S** is twice this area.

If we're still thinking in rectangular coordinates, we might *want* to integrate * ScscÎ¸dÎ¸*. THIS IS WRONG! (

__But if you're interested...__) But remember that when we integrate

*, it's because we're adding up the areas of a bunch of*

**f(x)dx***with one side-length*

**rectangles***and the other*

**f(x)***.*

**dx**For polar coordinates, we want to add up the areas of ** circles**! More specifically,

**of circles. Remember that if we have a slice of circle of radius**

*sectors***made by an angle**

*r***to the x-axis, then that sector has area (**

*Î¸***So the expression we really want to integrate given the expression**

*Î¸/2)(r^2).***is**

*r = f(Î¸)***.**

*(1/2)(f(Î¸)^2)dÎ¸*Back to our function ** r = ScscÎ¸dÎ¸**, we find that the area of a square of side-length

**is**

*S*Let's do it!

Remember that the derivative of ** cotÎ¸** is

**, since**

*-(cscÎ¸)^2*So we finally find the area of a square with side-length ** S** to be...

**Thank you for reading!**

**Jonathan Gerhard**

**J****Math****G**

**J**

**Math**

**G**