Calculus – Multiple integrals – Double integrals in polar coordinates

For regions which are part of a disk, we might use polar coordinates: \(\left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm]y=r\sin(\theta)\end{array}\right.\quad\Longrightarrow\quad x^2+y^2=r^2\).

We start with a polar rectangle

\[R=\{(r\cos(\theta),r\sin(\theta))\,|\,a\leq r\leq b,\;\alpha\leq\theta\leq\beta\}\quad\textrm{with}\quad a\geq0\quad\textrm{and}\quad\beta-\alpha\leq2\pi.\]

In order to find the double integral \(\displaystyle\iint\limits_Rf(x,y)\,dA\) over such a polar rectangle \(R\), we divide the interval \([a,b]\) into \(m\) subintervals \([r_{i-1},r_i]\) with equal width \(\Delta r=(b-a)/m\) and we divide the interval \([\alpha,\beta]\) into \(n\) subintervals \([\theta_{j-1},\theta_j]\) of equal width \(\Delta\theta=(\beta-\alpha)/n\).

Now we consider the polar rectangles

\[R_{ij}=\{(r\cos(\theta),r\sin(\theta))\,|\,r_{i-1}\leq r\leq r_i,\;\theta_{j-1}\leq\theta\leq\theta_j\}\]

with area (let: \(r_i^*=(r_i+r_{i-1})/2\))

\[\Delta A_{ij}=\frac{1}{2}r_i^2\,\Delta\theta-\frac{1}{2}r_{i-1}^2\,\Delta\theta=\frac{1}{2}(r_i^2-r_{i-1}^2)\,\Delta\theta =\frac{1}{2}(r_i+r_{i-1})(r_i-r_{i-1})\,\Delta\theta=r_i^*\,\Delta r\,\Delta\theta.\]

Here we have: \(\Delta r=r_i-r_{i-1}\) and \(\Delta\theta=\theta_j-\theta_{j-1}\).

It is easier to approximate the area of a polar rectangle. Note that the length of an arc of a circle with radius \(r\) and difference of angles \(\Delta\theta\) equals \(r\,\Delta\theta\). This implies that the area of a polar rectangle is approximately equal to the area of a rectangle with sides \(\Delta r\) and \(r\,\Delta\theta\), which is: \(r\,\Delta r\,\Delta\theta\).

Theorem: If \(f\) is continuous on a polar rectangle \(R\) given by \(0\leq a\leq r\leq b\) and \(\alpha\leq\theta\leq\beta\) with \(\beta-\alpha\leq2\pi\), then

\[\iint\limits_Rf(x,y)\,dA=\int_{\alpha}^{\beta}\int_a^bf(r\cos(\theta),r\sin(\theta))\,r\,dr\,d\theta.\]

In polar coordinates \(\left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm]y=r\sin(\theta)\end{array}\right.\) the area element equals \(dA=r\,dr\,d\theta\).

Examples:

1) Stewart §15.3, Example 1
Evaluate \(\displaystyle\iint\limits_R(3x+4y^2)\,dA\), where \(D\) is the region in the upper half-plane bounded by the circles \(x^2+y^2=1\) and \(x^2+y^2=4\).

Solution: Note that the region \(R=\{(x,y)\,|\,y\geq0,\;1\leq x^2+y^2\leq 4\}\) is described in polar coordinates as

\[R=\{(r\cos(\theta),r\sin(\theta))\,|\,1\leq r\leq 2,\;0\leq\theta\leq\pi\}.\]

Hence we have using \(\cos(2\theta)=1-2\sin^2(\theta)\):

\begin{align*} \iint\limits_R(3x+4y^2)\,dA&=\int_0^{\pi}\int_1^2\left(3r\cos(\theta)+4r^2\sin^2(\theta)\right)\,r\,dr\,d\theta =\int_0^{\pi}\bigg[r^3\cos(\theta)+r^4\sin^2(\theta)\bigg]_{r=1}^2\,d\theta\\[2.5mm] &=\int_0^{\pi}\left(7\cos(\theta)+15\sin^2(\theta)\right)\,d\theta =\int_0^{\pi}\left(7\cos(\theta)+\frac{15}{2}\left(1-\cos(2\theta)\right)\right)\,d\theta\\[2.5mm] &=\bigg[7\sin(\theta)+\frac{15}{2}\theta-\frac{15}{4}\sin(2\theta)\bigg]_{\theta=0}^{\pi}=\frac{15}{2}\pi. \end{align*}

2) Stewart §15.3, Example 2
Find the volume of the solid bounded by the plane \(z=0\) and the paraboloid \(z=1-x^2-y^2\).

Solution: Note that the intersection of the paraboloid and the \(xy\)-plane (\(z=0\) is the circle \(x^2+y^2=1\). So, the solid lies under the paraboloid and above the circular disk \(D\) given by \(x^2+y^2\leq1\). In polar coordinates \(D\) is given by

\[D=\{(r\cos(\theta),r\sin(\theta))\,|\,0\leq r\leq1,\;0\leq\theta\leq2\pi\}.\]

Hence the volume equals

\[V=\iint\limits_D(1-x^2-y^2)\,dA=\int_0^{2\pi}\int_0^2(1-r^2)\,r\,dr\,d\theta=\int_0^{2\pi}d\theta\,\int_0^1(r-r^3)\,dr =2\pi\bigg[\frac{1}{2}r^2-\frac{1}{4}r^4\bigg]_{r=0}^1=\frac{1}{2}\pi.\]

More general regions

Theorem: If \(f\) is continuous on a polar region of the form

\[D=\{(r\cos(\theta),r\sin(\theta))\,|\,\alpha\leq\theta\leq\beta,\;h_1(\theta)\leq r\leq h_2(\theta)\}\quad\textrm{with}\quad\beta-\alpha\leq2\pi,\]

then

\[\iint\limits_Df(x,y)\,dA=\int_{\alpha}^{\beta}\int_{h_1(\theta)}^{h_2(\theta)}f(r\cos(\theta),r\sin(\theta))\,r\,dr\,d\theta.\]

Examples:

1) Stewart §15.3, Example 3
Use a double integral to find the area of the region \(D\) enclosed by one loop of the four-leaved rose \(r=\cos(2\theta)\).

Solution: Note that the region \(D\) in polar coordinates is given by

\[D=\{(r\cos(\theta),r\sin(\theta))\,|\,-\textstyle\frac{1}{4}\pi\leq\theta\leq\frac{1}{4}\pi,\;0\leq\theta\leq\cos(2\theta)\}.\]

Hence we have using \(\cos(2x)=2\cos^2(x)-1\):

\begin{align*} \textrm{area}(D)&=\iint\limits_D1\,dA=\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\int_0^{\cos(2\theta)}r\,dr\,d\theta =\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\bigg[\frac{1}{2}r^2\bigg]_{r=0}^{\cos(2\theta)}\,d\theta =\frac{1}{2}\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\cos^2(2\theta)\,d\theta\\[2.5mm] &=\frac{1}{4}\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\left(1+\cos(4\theta)\right)\,d\theta =\frac{1}{4}\bigg[\theta+\frac{1}{4}\sin(4\theta)\bigg]_{\theta=-\frac{1}{4}\pi}^{\frac{1}{4}\pi}=\frac{1}{8}\pi. \end{align*}

2) Stewart §15.3, Example 4
Find the volume of the solid that lies under the paraboloid \(z=x^2+y^2\), above the \(xy\)-plane and inside the cylinder \(x^2+y^2=2x\).

Solution: The solid lies above the disk \(D\) with boundary circle given by \(x^2+y^2=2x\) or equivalently, completing the square, by \((x-1)^2+y^2=1\). In polar coordinates we have \(x^2+y^2=r^2\) and \(x=r\cos(\theta)\), so this boundary circle becomes \(r^2=2r\cos(\theta)\) or \(r=2\cos(\theta)\). Hence we have:

\[D=\{(r\cos(\theta),r\sin(\theta))\,|\,-\textstyle\frac{1}{2}\pi\leq\theta\leq\frac{1}{2}\pi,\;0\leq r\leq2\cos(\theta)\}.\]

Hence the volume equals using \(\cos(2\theta)=2\cos^2(\theta)-1\)

\begin{align*} V&=\iint\limits_D(x^2+y^2)\,dA=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\int_0^{2\cos(\theta)}r^2\,r\,dr\,d\theta =\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\bigg[\frac{1}{4}r^4\bigg]_{r=0}^{2\cos(\theta)}\,d\theta =4\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\cos^4(\theta)\,d\theta=8\int_0^{\frac{1}{2}\pi}\cos^4(\theta)\,d\theta\\[2.5mm] &=8\int_0^{\frac{1}{2}\pi}\left(\frac{1+\cos(2\theta)}{2}\right)^2\,d\theta =2\int_0^{\frac{1}{2}\pi}\left(1+2\cos(2\theta)+\cos^2(2\theta)\right)\,d\theta =2\int_0^{\frac{1}{2}\pi}\left(1+2\cos(2\theta)+\frac{1}{2}\left(1+\cos(4\theta)\right)\right)\,d\theta\\[2.5mm] &=\bigg[3\theta+2\sin(2\theta)+\frac{1}{2}\sin(4\theta)\bigg]_{\theta=0}^{\frac{1}{2}\pi}=\frac{3}{2}\pi. \end{align*}

Applications

1) Stewart §15.3, Exercise 35
A swimming pool is circular with a diameter of \(10\) m. The depth is constant along east-west lines and increases linearly from \(1\) m at the south end to \(2\) m at the north end. Find the volume of water in the pool.

Solution: Note that the linear function \(f(x)\) such that \(f(-5)=1\) and \(f(5)=2\) is given by \(f(x)=\displaystyle\frac{1}{10}x+\frac{3}{2}=\frac{x+15}{10}\). Then the volume is given by \(V=\displaystyle\iint\limits_{x^2+y^2\leq25}\frac{x+15}{10}\,dA\).

Using polar coordinates we obtain

\[V=\int_0^{2\pi}\int_0^5\frac{r\cos(\theta)+15}{10}\,r\,dr\,d\theta =\int_0^{2\pi}\bigg[\frac{1}{30}r^3\cos(\theta)+\frac{3}{4}r^2\bigg]_{r=0}^5\,d\theta =\int_0^{2\pi}\left(\frac{25}{6}\cos(\theta)+\frac{75}{4}\right)\,d\theta=0+\frac{75}{4}\cdot2\pi=\frac{75}{2}\pi.\]

2) Proof of the Gaussian integral
Earlier we have seen that the Gaussian integral is: \(\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}\). Now we are able to prove this result.

Proof: We start with \(\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\,dx=2\int_0^{\infty}e^{-x^2}\,dx\) and \(\displaystyle\int_0^{\infty}e^{-x^2}\,dx=\lim\limits_{R\to\infty}\int_0^Re^{-x^2}\,dx\). Now we have

\[\left(\int_0^Re^{-x^2}\,dx\right)^2=\left(\int_0^Re^{-x^2}\,dx\right)\left(\int_0^Re^{-y^2}\,dy\right) =\iint\limits_{[0,R]\times[0,R]}e^{-x^2-y^2}\,dA.\]

Note that

\[\iint\limits_{C_1}e^{-x^2-y^2}\,dA\leq\iint\limits_{[0,R]\times[0,R]}e^{-x^2-y^2}\,dA\leq\iint\limits_{C_2}e^{-x^2-y^2}\,dA\]

with \(C_1=\{(x,y)\,|\,x\geq0,\;y\geq0,\;x^2+y^2\leq R^2\}\) and \(C_2=\{(x,y)\,|\,x\geq0,\;y\geq0,\;x^2+y^2\leq 2R^2\}\).

Now we use polar coordinates to obtain

\[\iint\limits_{C_1}e^{-x^2-y^2}\,dA=\int_0^{\frac{1}{2}\pi}\int_0^Re^{-r^2}\,r\,dr\,d\theta =\frac{1}{2}\pi\bigg[-\frac{1}{2}e^{-r^2}\bigg]_{r=0}^R=\frac{1}{4}\pi\left(1-e^{-R^2}\right)\]

and

\[\iint\limits_{C_2}e^{-x^2-y^2}\,dA=\int_0^{\frac{1}{2}\pi}\int_0^{R\sqrt{2}}e^{-r^2}\,r\,dr\,d\theta =\frac{1}{2}\pi\bigg[-\frac{1}{2}e^{-r^2}\bigg]_{r=0}^{R\sqrt{2}}=\frac{1}{4}\pi\left(1-e^{-2R^2}\right).\]

Finally we take the limit \(R\to\infty\) and use the squeeze theorem to obtain that \(\displaystyle\lim\limits_{R\to\infty}\iint\limits_{[0,R]\times[0,R]}e^{-x^2-y^2}\,dA=\frac{1}{4}\pi\). Hence we have

\[\left(\int_0^{\infty}e^{-x^2}\,dx\right)^2=\lim\limits_{R\to\infty}\left(\int_0^Re^{-x^2}\,dx\right)^2=\frac{1}{4}\pi,\]

which implies that \(\displaystyle\int_0^{\infty}e^{-x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}\). This proves that \(\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}\).

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Last modified on September 28, 2021