Calculus – Multiple integrals – Triple integrals in cylindrical coordinates

Cylindrical coordinates are:

\[\left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm]y=r\sin(\theta)\\[2.5mm]z=z\end{array}\right.\quad\Longrightarrow\quad r^2=x^2+y^2\quad\textrm{and}\quad\tan(\theta)=\frac{y}{x}.\]

The volume element equals the area of the polar rectangle multiplied by the height: \(dV=r\,dz\,dr\,d\theta\).

Suppose that \(E\) is a type 1 region whose projection \(D\) onto the \(xy\)-plane is conveniently described in polar coordinates. In particular, suppose that \(f\) is continuous and

\[E=\{(x,y,z)\,|\,(x,y)\in D,\;u_1(x,y)\leq z\leq u_2(x,y)\},\]

where \(D\) is given in polar coordinates by

\[D=\{(r\cos(\theta),r\sin(\theta))\,|\,\alpha\leq\theta\leq\beta,\;h_1(\theta)\leq r\leq h_2(\theta)\}.\]

Then we have

\[\iiint\limits_Ef(x,y,z)\,dV=\int_{\alpha}^{\beta}\int_{h_1(\theta)}^{h_2(\theta)}\int_{u_1(r\cos(\theta),r\sin(\theta))}^{u_2(r\cos(\theta),r\sin(\theta))} f(r\cos(\theta),r\sin(\theta),z)\,r\,dz\,dr\,d\theta.\]

Stewart §15.7, Example 3
A solid \(E\) lies within the cylinder \(x^2+y^2=1\), below the plane \(z=4\), and above the paraboloid \(z=1-x^2-y^2\). The mass density at any point is proportional to its distance from the axis of the cylinder. Find the mass of \(E\).

Solution: In cylindrical coordinates we have

\[E=\{(r\cos(\theta),r\sin(\theta),z)\,|\,0\leq\theta\leq2\pi,\;0\leq r\leq1,\;1-r^2\leq z\leq 3\}.\]

Since the mass density at \((x,y,z)\) is proportional to the distance from the \(z\)-axis, the mass density function is \(K\sqrt{x^2+y^2}=K\,r\) with \(K\) a constant. Hence we have:

\begin{align*} m&=\iiint\limits_EK\sqrt{x^2+y^2}\,dV=\int_0^{2\pi}\int_0^1\int_{1-r^2}^4K\,r\,r\,dz\,dr\,d\theta=K\int_0^{2\pi}d\theta \int_0^1r^2\left(4-(1-r^2)\right)\,dr\\[2.5mm] &=K\,2\pi\int_0^1(3r^2+r^4)\,dr=2\pi\,K\bigg[r^3+\frac{1}{5}r^5\bigg]_{r=0}^1=\frac{12}{5}\pi\,K. \end{align*}

Stewart §15.7, Example 4
Evaluate \(\displaystyle\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{x^2+y^2}}^2(x^2+y^2)\,dz\,dy\,dx\).

Solution: This iterated integral is a triple integral over the solid region

\[E=\{(x,y,z)\,|\,-2\leq x\leq2,\;-\sqrt{4-x^2}\leq y\leq\sqrt{4-x^2},\;\sqrt{x^2+y^2}\leq z\leq 2\}\]

and the projection of \(E\) onto the \(xy\)-plane is the disk \(x^2+y^2\leq4\). The lower surface of \(E\) is the cone \(z=\sqrt{x^2+y^2}\) and its upper surface is the plane \(z=4\). So in cylindrical coordinates we have

\[E=\{(r\cos(\theta),r\sin(\theta),z)\,|\,0\leq\theta\leq2\pi,\;0\leq r\leq 2,\;r\leq z\leq2\}.\]

Therefore we have

\begin{align*} \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{x^2+y^2}}^2(x^2+y^2)\,dz\,dy\,dx&=\iiint\limits_E(x^2+y^2)\,dV =\int_0^{2\pi}\int_0^2\int_r^2r^2\,r\,dz\,dr\,d\theta=\int_0^{2\pi}d\theta\int_0^2r^3(2-r)\,dr\\[2.5mm] &=2\pi\bigg[\frac{1}{2}r^4-\frac{1}{5}r^5\bigg]_{r=0}^2=2\pi\left(8-\frac{32}{5}\right)=2\pi\cdot\frac{8}{5}=\frac{16}{5}\pi. \end{align*}

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Last modified on October 4, 2021