By the end of this section, you should be able to answer the following questions:

• What is the relationship between polar coordinates and rectangular coordinates?
• How do you transform a double integral in rectangular coordinates into one in terms of polar coordinates?
• What is the Jacobian of the transformation?

For polar coordinates, we have \[ x = r\cos \theta, \quad y = r \sin \theta. \]

Consider the volume of a solid beneath a surface $z = f(x,y)$ and above a circular region in the $x$-$y$ plane.

Therefore the volume under the surface and above each polar rectangle can be approximated as \[ \text{vol. on box} \approx r^*~ \Delta \theta ~\Delta r ~f\left( r^*\cos \theta^*, r^* \sin \theta^* \right). \]

Here $f\left( r^*\cos \theta^*, r^* \sin \theta^* \right)$ is the value of the function at the point $P$, which is also the height of the box used in the approximation.

To obtain an approximation for the entire volume below the surface, we sum over the entire polar grid:

The double integral in rectangular coordinates is then transformed as follows: \[ \iint_\limits{R} f(x,y)~dx~dy = \iint_\limits{S} f\left( r \cos \theta , r \sin \theta \right)r ~dr~d\theta. \]

🤔 Unfortunately, we cannot evaluate this integral in rectangular coordinates.

We need to use polar coordinates: $x= r\cos\theta,$ $y=r\sin \theta$.

This implies that $x^2+y^2 = r^2$. So \[ \iint_\limits{D}e^{-\left(x^2 + y^2 \right)}~dx~dy = \iint_\limits{D} e^{-r^2} r ~dr~d\theta. \]

The region $D$ can be described in polar coordinates. That is: $$0\leq r \leq R \;\text{ and } \;0 \leq \theta \leq 2\pi.$$

Then $D = \left\{ \left( r, \theta \right) ~|~ 0\leq r \leq R,0 \leq \theta \leq 2\pi \right\}$.

😃 We can evaluate this integral in rectangular and polar coordinates.

Let's use polar coordinates: $x= r\cos\theta,$ $y=r \sin \theta.$

This implies that our surface is defined as $1-x^2-y^2 = 1-r^2,$ and the domain $D$ is region inside the circle \[ x^2+y^2=1 \] which is determined by intersecting $z=0$ and $z=1-x^2-y^2.$

Thus we have that $$0\leq r \leq 1 \;\text{ and } \;0 \leq \theta \leq 2\pi$$

So $D = \left\{ \left( r, \theta \right) ~|~ 0\leq r \leq 1,0 \leq \theta \leq 2\pi \right\}$.

First, we need to find the boundary.

If we complete the square in $x^2+y^2=2x,$ we obtain \[ (x-1)^2+y^2=1. \]

This is the cylinder centred at $(1,0)$ with radius $1$. In this example the domain is defined by the expression \[ x^2+y^2=2x. \]

Again it is simpler to use polar coordinates.

Thus, substituing the values $\,x= r\cos\theta\,$ and $\,y=r\sin \theta\,$ in $$x^2+y^2=2x,$$ we get $r^2\cos ^2\theta +r^2\sin ^2\theta = 2r \cos \theta$. That is $$ r^2 = 2r \cos \theta \Rightarrow r = 2 \cos \theta \qquad (r\neq 0).$$

Thus $$D = \left\{ \left( r, \theta \right) ~|~ 0\leq r \leq 2\cos \theta, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right\}.$$

$\text{V}$ $\displaystyle =\iint_{D} \left(x^2+y^2 \right)dA $ $\displaystyle= \iint_{D} r^2r ~dr~d\theta $

$\;\,\, \displaystyle =\int_{\theta=-\frac{\pi}{2}}^{\theta = \frac{\pi}{2}} \int_{r=0}^{r=2\cos \theta} r^3 ~dr~d\theta $

$\;\,\,\displaystyle =\int_{\theta=-\frac{\pi}{2}}^{\theta = \frac{\pi}{2}} \left[ \frac{r^4}{4} \Bigg|_{r=0}^{r=2\cos \theta} \right] d\theta $ $\displaystyle = \int_{\theta=-\frac{\pi}{2}}^{\theta = \frac{\pi}{2}} 4\cos^4 \theta~ d\theta $

$\;\,\,\displaystyle = \int_{\theta=-\frac{\pi}{2}}^{\theta = \frac{\pi}{2}} \left( \frac{3}{2} + 2 \cos \left(2 \theta\right) + \frac{1}{2} \cos \left(4 \theta\right) \right) d\theta $

$\;\,\,\displaystyle = \left[\frac{3}{2}\theta +\sin\left(2 \theta\right)+\frac{1}{8}\sin\left(4 \theta\right)\right] \Bigg|_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}$ $\displaystyle = \large \frac{3\pi}{2} .\qquad \qquad\qquad$