How to calculate area of circle $(x-p)^2+(y-q)^2=r^2$ using double integrals?
I am assuming that area should be equal to area of circle $x^2+y^2=r^2$ because $p$ and $q$ doesn't affect the dimensions of circle just Its position.
So if we calculate double integral over region $x^2+y^2=r^2$, using polar coordinates, we get:
$ \int_0^{2\pi} d\phi \int_0^r \rho d\rho = r^2\pi $
So is It same for equation $(x-p)^2+(y-q)^2=r^2$ ?
I have same question for volume of sphere.Using triple integrals to calculate Its volume, do we get different results for $x^2+y^2+z^2<=1$ and $x^2+(y-1)^2+(z-2)^2 <= 1$ ?
You mean this integral? $$ \int_{p-r}^{p+r} \left(\int_{q-\sqrt{r^2-(x-p)^2}}^{q+\sqrt{r^2-(x-p)^2}} \,dy \right)\,dx $$