Calculus – Multiple integrals – More applications 1

Examples

1) Consider the region \(R\) enclosed by the circles \(x^2+y^2=1\) and \(x^2+y^2=4\) and the lines \(y=\displaystyle\frac{x}{\sqrt{3}}\) and \(y=x\sqrt{3}\) in the first quadrant.

Using polar coordinates we obtain for the area of \(R\):

\[\textrm{area}(R)=\iint\limits_R1\,dA=\int_{\arctan\left(\frac{1}{\sqrt{3}}\right)}^{\arctan\left(\sqrt{3}\right)}\int_1^2r\,dr\,d\theta =\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}d\theta\int_1^2r\,dr=\left(\frac{1}{3}\pi-\frac{1}{6}\pi\right)\bigg[\frac{1}{2}r^2\bigg]_{r=1}^2 =\frac{1}{6}\pi\cdot\frac{3}{2}=\frac{1}{4}\pi.\]

Instead it is also possible to apply the change of variables \(u=x^2+y^2\) and \(v=\displaystyle\frac{y}{x}\) which transforms the region

\[R=\{(x,y)\,|\,1\leq x^2+y^2\leq4,\;0 < \displaystyle\frac{x}{\sqrt{3}}\leq y\leq x\sqrt{3}\}\]

in the first quadrant of the \(xy\)-plane into the rectangular region

\[S=\{(u,v)\,|\,1\leq u\leq 4,\;\displaystyle\frac{1}{\sqrt{3}}\leq v\leq\sqrt{3}\}\]

in the \(uv\)-plane. Note that

\[\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y}\\[2.5mm] \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y}\end{vmatrix} =\begin{vmatrix}2x & 2y\\[2.5mm]-\displaystyle\frac{y}{x^2} & \displaystyle\frac{1}{x}\end{vmatrix}=2+2\,\frac{y^2}{x^2}=2(1+v^2)\quad\Longrightarrow\quad \frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\displaystyle\frac{\partial(u,v)}{\partial(x,y)}}=\frac{1}{2(1+v^2)}.\]

This leads to

\begin{align*} \textrm{area}(R)&=\iint\limits_R1\,dA=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\int_1^4\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv =\frac{1}{2}\int_1^4du\,\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{dv}{1+v^2}\\[2.5mm] &=\frac{3}{2}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(\frac{1}{\sqrt{3}}\right)\right) =\frac{3}{2}\left(\frac{1}{3}\pi-\frac{1}{6}\pi\right)=\frac{3}{2}\cdot\frac{1}{6}\pi=\frac{1}{4}\pi. \end{align*}

2) Consider the region \(R\) enclosed by the hyperbolas \(xy=1\) and \(xy=2\) and the lines \(y=x\) and \(y=2x\) in the first quadrant.

The change of variables \(u=xy\) and \(v=\displaystyle\frac{y}{x}\) transforms the region \(R=\{(x,y)\,|\,1\leq xy\leq2,\;0 < x\leq y\leq2x\}\) in the \(xy\)-plane into the rectangular region \(S=\{(u,v)\,|\,1\leq u\leq 2,\;1\leq v\leq 2\}\) in the \(uv\)-plane. Note that

\[\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y}\\[2.5mm] \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y}\end{vmatrix} =\begin{vmatrix}y & x\\[2.5mm]-\displaystyle\frac{y}{x^2} & \displaystyle\frac{1}{x}\end{vmatrix}=\frac{y}{x}+\frac{y}{x}=2v \quad\Longrightarrow\quad\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\displaystyle\frac{\partial(u,v)}{\partial(x,y)}}=\frac{1}{2v}.\]

This leads to

\[\textrm{area}(R)=\iint\limits_R1\,dA=\int_1^2\int_1^2\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv= \int_1^2\int_1^2\frac{1}{2v}\,du\,dv=\frac{1}{2}\bigg[\ln(v)\bigg]_{v=1}^2=\frac{1}{2}\ln(2).\]

Note that the points of intersection of the hyperbolas and the lines are: \((\frac{1}{2}\sqrt{2},\sqrt{2})\), \((1,1)\), \((1,2)\) and \((\sqrt{2},\sqrt{2})\). So, in rectangular or Cartesian coordinates, we would obtain

\begin{align*} \textrm{area}(R)&=\iint\limits_E1\,dA=\int_{\frac{1}{2}\sqrt{2}}^1\int_{\frac{1}{x}}^{2x}dy\,dx+\int_1^{\sqrt{2}}\int_x^{\frac{2}{x}}dy\,dx =\int_{\frac{1}{2}\sqrt{2}}^1\left(2x-\frac{1}{x}\right)\,dx+\int_1^{\sqrt{2}}\left(\frac{2}{x}-x\right)\,dx\\[2.5mm] &=\bigg[x^2-\ln(x)\bigg]_{\frac{1}{2}\sqrt{2}}^1+\bigg[2\ln(x)-\frac{1}{2}x^2\bigg]_1^{\sqrt{2}} =1-\frac{1}{2}+\ln(\tfrac{1}{2}\sqrt{2})+2\ln(\sqrt{2})-1+\frac{1}{2}=\frac{1}{2}\ln(2). \end{align*}

3) Consider the region \(R\) bounded by the parabolas \(y=x^2\) and \(y=8x^2\) and the hyperbolas \(xy=1\) and \(xy=8\).

The change of variables \(u=xy\) and \(v=\displaystyle\frac{y}{x^2}\) transforms the region \(R=\{(x,y)\,|\,1\leq xy\leq8,\;x^2\leq y\leq8x^2\}\) in the \(xy\)-plane into the rectangular region \(S=\{(u,v)\,|\,1\leq u\leq 8,\;1\leq v\leq 8\}\) in the \(uv\)-plane. Note that

\[\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y}\\[2.5mm] \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y}\end{vmatrix} =\begin{vmatrix}y & x\\[2.5mm]-\displaystyle\frac{2y}{x^3} & \displaystyle\frac{1}{x^2}\end{vmatrix}=\frac{y}{x^2}+\frac{2y}{x^2}=3v \quad\Longrightarrow\quad\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\displaystyle\frac{\partial(u,v)}{\partial(x,y)}}=\frac{1}{3v}.\]

This leads to

\[\textrm{area}(R)=\iint\limits_R1\,dA=\int_1^8\int_1^8\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv= \int_1^8\int_1^8\frac{1}{3v}\,du\,dv=\frac{7}{3}\bigg[\ln(v)\bigg]_{v=1}^8=\frac{7}{3}\ln(8)=7\ln(2).\]

Note that the points of intersection of the hyperbolas and the lines are: \((\frac{1}{2},2)\), \((1,1)\), \((1,8)\) and \((2,4)\). So, in rectangular or Cartesian coordinates, we would obtain

\begin{align*} \textrm{area}(R)&=\iint\limits_E1\,dA=\int_{\frac{1}{2}}^1\int_{\frac{1}{x}}^{8x^2}dy\,dx+\int_1^2\int_{x^2}^{\frac{8}{x}}dy\,dx =\int_{\frac{1}{2}}^1\left(8x^2-\frac{1}{x}\right)\,dx+\int_1^2\left(\frac{8}{x}-x^2\right)\,dx\\[2.5mm] &=\bigg[\frac{8}{3}x^3-\ln(x)\bigg]_{\frac{1}{2}}^1+\bigg[8\ln(x)-\frac{1}{3}x^3\bigg]_1^2 =\frac{8}{3}-\frac{1}{3}+\ln(\tfrac{1}{2})+8\ln(2)-\frac{8}{3}+\frac{1}{3}=7\ln(2). \end{align*}

Proofs of the Basel problem

The Basel problem is to find the sum of the convergent series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}\). The problem is named after the hometown of the Swiss mathematician Leonhard Euler (1707-1783), who solved the problem in 1734. Earlier we have seen that this is a special case of the Riemann zeta function \(\displaystyle\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}\) for \(\textrm{Re}(s)>1\). We only consider real values of \(s\); in that case we know that the series is absolutely convergent if \(s>1\). However, in general it is not very easy to find its sum. The Basel problem is to find the sum of \(\zeta(2)\).

1) A possible proof is based on the observation that \(\displaystyle\int_0^1t^{n-1}\,dt=\frac{1}{n}\) for \(n>0\).

Then we have using the geometric series \(\displaystyle\sum_{n=1}^{\infty}t^{n-1}=\frac{1}{1-t}\) for \(|t|<1\):

\[\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_0^1\int_0^1\sum_{n=1}^{\infty}(xy)^{n-1}\,dx\,dy =\int_0^1\int_0^1\frac{dx\,dy}{1-xy}.\]

In order to evaluate this iterated integral, we apply the substitution \(u=(x+y)/2\) and \(y=(x-y)/2\) or \(x=u-v\) and \(y=u+v\) with Jacobian

\[\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix} =\begin{vmatrix}1 & -1\\[2.5mm]1 & 1\end{vmatrix}=1+1=2,\]

to obtain that

\[\int_0^1\int_0^1\frac{dx\,dy}{1-xy}=\iint\limits_S\frac{1}{1-(u-v)(u+v)}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv =2\iint\limits_S\frac{du\,dv}{1-u^2+v^2},\]

where \(S\) is the square with vertices \((0,0)\), \((\frac{1}{2},-\frac{1}{2})\), \((1,0)\) and \((\frac{1}{2},\frac{1}{2})\):

Using the symmetry we obtain that

\begin{align*} \zeta(2)&=2\iint\limits_S\frac{du\,dv}{1-u^2+v^2}=4\int_0^{\frac{1}{2}}\int_0^u\frac{dv\,du}{1-u^2+v^2} +4\int_{\frac{1}{2}}^1\int_0^{1-u}\frac{dv\,du}{1-u^2+v^2}\\[2.5mm] &=4\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\,du +4\int_{\frac{1}{2}}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\,du. \end{align*}

Observe that

\[\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)=\arctan(u)\quad\textrm{and}\quad2\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)=\arccos(u), \quad -1 < u < 1.\]

This implies that

\begin{align*} \zeta(2)&=4\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\,du +4\int_{\frac{1}{2}}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\,du\\[2.5mm] &=4\int_0^{\frac{1}{2}}\frac{\arcsin(u)}{\sqrt{1-u^2}}\,du+2\int_{\frac{1}{2}}^1\frac{\arccos(u)}{\sqrt{1-u^2}}\,du =2\arcsin^2(u)\bigg|_0^{\frac{1}{2}}-\arccos^2(u)\bigg|_{\frac{1}{2}}^1\\[2.5mm] &=2\left(\frac{1}{6}\pi\right)^2+\left(\frac{1}{3}\pi\right)^2=\frac{1}{18}\pi^2+\frac{1}{9}\pi^2=\frac{1}{6}\pi^2. \end{align*}

2) Another proof is based on the observation that

\[\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}=\sum_{n=1}^{\infty}\frac{1}{n^s}-\sum_{n=1}^{\infty}\frac{1}{(2n)^s} =\left(1-\frac{1}{2^s}\right)\zeta(s).\]

Observe that \(\displaystyle\int_0^1x^{2n}\,dx=\frac{1}{2n+1}\) for \(n>0\).

Hence we have using the geometric series \(\displaystyle\sum_{n=0}^{\infty}t^n=\frac{1}{1-t}\) for \(|t|<1\)

\[\frac{3}{4}\zeta(2)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\int_0^1\int_0^1\sum_{n=0}^{\infty}(xy)^{2n}\,dx\,dy =\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}.\]

Applying the change of variables \(x=\sin(u)/\cos(v)\) and \(y=\sin(v)/\cos(u)\) with Jacobian

\[\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix} =\begin{vmatrix}\displaystyle\frac{\cos(u)}{\cos(v)} & \displaystyle\frac{\sin(u)\sin(v)}{\cos^2(v)}\\[2.5mm] \displaystyle\frac{\sin(u)\sin(v)}{\cos^2(u)} & \displaystyle\frac{\cos(v)}{\cos(u)}\end{vmatrix} =1-\frac{\sin^2(u)\sin^2(v)}{\cos^2(u)\cos^2(v)}=1-x^2y^2,\]

we obtain

\[\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}=\iint\limits_Tdu\,dv,\]

where \(T=\{(u,v)\,|\,u>0,\;v>0,\;u+v=\pi/2\}\):

Hence we have

\[\frac{3}{4}\zeta(2)=\iint\limits_Tdu\,dv=\textrm{area}(T)=\frac{1}{2}\cdot\frac{1}{4}\pi^2=\frac{1}{8}\pi^2 \quad\Longrightarrow\quad\zeta(2)=\frac{4}{3}\cdot\frac{1}{8}\pi^2=\frac{1}{6}\pi^2.\]

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Last modified on October 6, 2021