In the double integral double integral over D z(x,y) dx

In the double integral $\iint_D z(x,y)\,dx\,dy$, convert to polar coordinates $\varphi$ and $r$ and write the limits of integration for both possible orders. The region $D$ is the circle $x^2+y^2<ax$ where $a>0$.

Step 1 · Rewrite the circle in standard form

$$ x^2-ax+y^2<0 \ \Longrightarrow\ \left(x-\frac a2\right)^2+y^2<\frac{a^2}{4}. $$

This is a circle centered at $\left(\frac a2,0\right)$ with radius $\frac a2$.

Step 2 · Convert to polar coordinates

With $x^2+y^2=r^2$ and $ax=ar\cos\varphi$, the inequality $x^2+y^2<ax$ becomes $r^2<ar\cos\varphi$, and for $r>0$:

$$ r < a\cos\varphi. $$

Step 3 · Determine the range of $\varphi$

For $r<a\cos\varphi$ to allow $r>0$ we need $\cos\varphi>0$, i.e. $-\frac{\pi}{2}<\varphi<\frac{\pi}{2}$.

Step 4 · Order 1 — integrate $r$ first, then $\varphi$

$$ \iint_D z\,dx\,dy = \int_{-\pi/2}^{\pi/2} d\varphi \int_0^{a\cos\varphi} z(r\cos\varphi,r\sin\varphi)\,r\,dr. $$

Step 5 · Order 2 — integrate $\varphi$ first, then $r$

The maximum $r$ on the region is $r_{\max}=a\cos(0)=a$ (the diameter), so $0<r<a$. From $r<a\cos\varphi$: $\cos\varphi>\frac ra$, so with $\varphi_0=\arccos\!\left(\frac ra\right)$:

$$ -\arccos\!\left(\frac ra\right) < \varphi < \arccos\!\left(\frac ra\right). $$

$$ \iint_D z\,dx\,dy = \int_0^{a} dr \int_{-\arccos(r/a)}^{\arccos(r/a)} z(r\cos\varphi,r\sin\varphi)\,r\,d\varphi. $$

Additional Notes

• Whenever a circle passes through the origin, its polar equation takes the simple form $r=2R_0\cos(\varphi-\varphi_0)$, where $R_0$ is the radius and $\varphi_0$ the angle to the center; here $\varphi_0=0$ and $R_0=\frac a2$, giving $r=a\cos\varphi$.