What have I calculated? Integrals

Question

I've got $$D = \{(x,y) \in R^2 : (x-1)^2+y^2<=1, (x-2)^2+y^2<=4\}$$ Which basically are two basically two circles. I tried to calculate the area of the $$(x-1)^2+y^2<=1$$ Using polar-coordinates: $$x = rcos\phi , y=rsin\phi$$

$$\int_{1}^{2}dr \int_{0}^{2\pi}((rcos\phi-1)^2+r^2sin^2\phi-1)d\phi$$ and I got $$\frac{14}{3}\pi$$

But, what did I actually calculated?

Answer 1

You shall use the substitution :

$$ dxdy \sim rdrd\theta $$ instead of $$ drd\theta $$ that you have used.

Answer 2

You calculate the volume below the following surface:

$$ \gamma(x,y)=(x-1)^2+y^2-1$$

For understanding this you can think that you use two parameters for your integral your resulting function (here $\gamma$) will generate a surface in 3D.