# How do transfer from a xy coordinate system to r and θ

Let S $$\subset$$ $$R^2$$ and let S be formed by :

• the region inside the circle $$x^2$$+$$y^2$$=9

• below the line y=x

• above the x axis

• laying to the right of x=1

Evaluate $$\int xydA$$.

I know how the region looks like.

The only problem I am having is how to draw the S region on a r and θ coordinate system.

Any help would be appreciated it.

EDIT:

The way to transform Cartesian to polar is to use a paramrtrization in terms of the radius of the circle and the angle. Meaning:

$$y = r \sin(\theta)$$ $$x = r \cos(\theta)$$

Which is such that:

$$x^2 + y^2 = r^2$$

Using the formulae for changing of variables, we have also to multiply the Jacobian for the polar coordinates, which happens to be $$J(x,y) = r$$ (the proof is quite simple).

You are integrating the region in first quadrant, in the intersection $$x=y$$ and the circumference starting from $$x=1$$.

Since $$x = r \cos(\theta)$$

For $$x=1$$ we have:

$$1 = r \cos(\theta) \Leftrightarrow r = \frac{1}{\cos(\theta)} = \sec(\theta)$$

Then for your particular case the region of integration $$A$$ is such that:

$$\sec(\theta) \leq r \leq 3$$ $$0 \leq \theta \leq \frac{\pi}{4}$$

Because the angle of $$y=x$$ is $$\pi/4$$. Then just integrate it:

$$\iint_{A} r^3 \cos(\theta)\sin(\theta) dr d\theta$$

• I have attached a picture of how the S region looks like Oct 12 '18 at 2:18
• Do you have to integrate it in polar coordinates? This is a vertically simple region, I think it's easier to integrate in Cartesian. Oct 12 '18 at 2:26
• Thank you so very much!!!! Oct 12 '18 at 2:37

You know how to draw it because the region is the same. The problem is to find equations for the boundaries. We have $$x=r\cos \theta$$ so you have that one. The circle is $$r=3.$$ Can you do the others?