21. Multiple Integrals in Curvilinear Coordinates

b. Integrating in Cylindrical Coordinates

4. Converting & Mixing, Rectangular & Cylindrical Coordinates

Sometimes it is useful to use cylindrical coordinates instead of rectangular coordinates when computing an integral. Other times it is best to use rectangular coordinates for one part of an integral and cylindrical coordinates for another part.

Converting Rectangular to Cylindrical Coordinates

Sometimes the antiderivative of the integrand is not known, making the integral difficult if not impossible. By converting to cylindrical coordinates, we can simplify the integral greatly.

Compute the integral \(\displaystyle I =\int_0^{3/\sqrt{2}}\int_y^{\sqrt{9-y^2}}\int_0^{x^2+y^2} \cos(z)\,dz\,dx\,dy\)

We now convert to cylindrical coordinates. We replace the volume differential by: \[ dV=dz\,dx\,dy=r\,dz\,dr\,d\theta \] and the new limits are: \[ 0 \le \theta \le \dfrac{\pi}{4} \qquad 0 \le r \le 3 \quad \text{and} \quad 0 \le z \le r^2 \] So the integral becomes \[\begin{aligned} I&=\int_0^{3/\sqrt{2}}\int_y^{\sqrt{9-y^2}}\int_0^{x^2+y^2} \cos(z)\,dz\,dx\,dy =\int_0^{\pi/4}\int_0^{3}\int_0^{r^2} \cos(z)\,r\,dz\,dr\,d\theta \\ &=\dfrac{\pi}{4}\int_0^{3} \left[\sin(z)\dfrac{}{}\right]_0^{r^2}\,r\,dr =\dfrac{\pi}{4}\int_0^{3} \sin(r^2)\,r\,dr =\dfrac{\pi}{4}\left[-\,\dfrac{1}{2}\,\cos(r^2)\right]_0^3 \\ &=\dfrac{\pi}{8}(-\cos(9)+\cos(0)) =\dfrac{\pi}{8}(1-\cos(9)) \end{aligned}\]

Mixing Rectangular and Cylindrical Coordinates

Sometimes one integral should be done in rectangular coordinates and another should be done in cylindrical coordinates, such as in the example below.

The cube \(-3 \le x \le 3\) and \(-3 \le y \le 3\) and \(6 \le z \le 12\) is cut out of the cone \(0 \le z \le 12-r\). Find the mass of the remaining part of the cone if the mass density is \(\delta=x^2+y^2=r^2\).

We compute the mass of the cone including the cube and then subtract the mass of the cube.

Using cylindrical coordinates, the mass of the cone including the cube is: \[\begin{aligned} M_\text{cone} &=\iiint\limits_\text{cone} \delta \,dV =\int_0^{2\pi}\int_0^{12}\int_0^{12-r} r^2\cdot r\,dz\,dr\,d\theta \\ &=2\pi\int_0^{12} r^3\left[z\dfrac{}{}\right]_{z=0}^{12-r}\,dr =2\pi\int_0^{12} r^3(12-r)\,dr \\ &=2\pi\left[3r^4-\dfrac{r^5}{5}\right]_0^{12} =2\pi\left(3\cdot12^4-\dfrac{12^5}{5}\right) \\ &=2\pi12^4\left(3-\dfrac{12}{5}\right) =\dfrac{6\cdot12^4\pi}{5} \end{aligned}\] Using rectangular coordinates, the mass of the cube is: \[\begin{aligned} M_\text{cube} &=\iiint\limits_\text{cube} \delta \,dV =\int_6^{12}\int_{-3}^3\int_{-3}^3 (x^2+y^2)\,dx\,dy\,dz \\ &=\int_6^{12} \,dz \left(\int_{-3}^3 x^2\,dx\int_{-3}^3 \,dy +\int_{-3}^3 \,dx\int_{-3}^3 y^2\,dy\right) \\ &=\left[\rule{0pt}{10pt}z\right]_6^{12} \left(\left[\dfrac{x^3}{3}\right]_{x=-3}^3\left[y\dfrac{}{}\right]_{y=-3}^3 +\left[x\dfrac{}{}\right]_{-3}^3\left[\dfrac{y^3}{3}\right]_{-3}^3\right) \\ &=6(18\cdot6+6\cdot18)=1296 \end{aligned}\] Thus, the mass inside the cone but outside the cube is \[ V=\dfrac{6\cdot12^4\pi}{5}-1296\approx76877. \]