Originally Posted by

**x3bnm** First of all thank you for taking the time and effort for explaining my question. The method you showed works for certain kind of integral. Say we have

$\displaystyle

\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy

$

Now:

$\displaystyle

x = \sqrt{1-y^2}\,\,\,\,\,\,\,\,\, (1)

$

$\displaystyle

x = -\sqrt{1-y^2}\,\,\,\,\,\,\,\,\,\ (2)

$

$\displaystyle

y = 1\,\,\,\,\,\,\,\,\,\ (3)

$

$\displaystyle

y = -1\,\,\,\,\,\,\,\,\,\ (4)

$

if we plug in $\displaystyle x =\rho \cos \varphi$ and $\displaystyle y =\rho \sin \varphi$ into equation (1) and solve for $\displaystyle \rho$ we get:

$\displaystyle

\rho = 1

$

Then plug in $\displaystyle y = \rho \sin \varphi$ and $\displaystyle \rho = 1$ into equation (3) and solve for $\displaystyle \varphi$ we get:

$\displaystyle

\varphi = \frac{\pi}{2}

$

Again if we plug in $\displaystyle y = \rho \sin \varphi$ and $\displaystyle \rho = 1$ into equation (4) and solve for $\displaystyle \varphi$ we get:

$\displaystyle

\varphi = \frac{3\pi}{2}

$

So the upper and lower bound of $\displaystyle \rho$ is $\displaystyle 1$ and $\displaystyle 0$ respectively and upper and

lower bound for $\displaystyle \varphi$ is $\displaystyle \frac{3\pi}{2}$ and $\displaystyle \frac{\pi}{2}$ respectively.

$\displaystyle

\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_{\frac{\pi}{2}}^\frac{3\pi}{2}\int_0^1\,\,r^3 \,\,d\rho\,\,d\varphi

$

But the answer is:

$\displaystyle

\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_0^{2\pi} \int_0^1\,\,r^3\,\,d\rho\,\,d\varphi

$

Why is the upper limit for $\displaystyle \varphi$ is $\displaystyle 2\pi$ instead of $\displaystyle \frac{3\pi}{2}$(what i found) and lower bound 0 instead of

$\displaystyle \frac{\pi}{2}$(what i found)? Why didn't the method you showed didn't work here? Did i make a mistake here?

Note: I edited this post couple of times to be correct. Sorry about that.