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6.4 Jacobina

  1. Evaluate \(\int_{S}\sqrt{x^{2}+y^{2}}dA\) where \(S=\{(x,y)\in \R^2:1\leq x^2+y^2\leq 2\}\)

    Annular region between two circles

    Let's use polar coordinates to describe the region and compute the integral.

    We use substitution \(x=r\cos \theta\) and \(y=r\sin\theta\)

    In fact we're using the map \(g:D\to S\) where \((r,\theta)\mapsto(x,y)=(r\cos\theta,r\sin\theta)\)

    We want \(g\) to be bijective. So \(\displaystyle \int_{S}fdA=\text{since f is continuous}=\iint_Sfdxdy=\iint _Df(g(r,\theta))\cdot |J_g(r,\theta)|drd\theta\)

    \(r\) represents the distance from the origin to the point

    \(\theta\) denote the angle from the positive x-axis, to the vector from the origin to the point

    In our case, \(0\leq \theta\leq 2\pi\) and \(1\leq r\leq \sqrt2\)

    imageimage​

    \(J_g = \begin{vmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{vmatrix} = \begin{vmatrix}\cos\theta & -r\sin\theta \\\sin\theta & r\cos\theta\end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r\)

    \(f(g(r,\theta)) = \sqrt{(r\cos\theta)^2 + (r\sin\theta)^2} = \sqrt{r^2\cos^2\theta + r^2\sin^2\theta} = \sqrt{r^2(\cos^2\theta+\sin^2\theta)} = \sqrt{r^2} = r\)

    Last we compute the integral \(\displaystyle \int_{S} \sqrt{x^{2}+y^{2}}\, dA = \int_{0}^{2\pi}\int_{1}^{\sqrt{2}}r \cdot r \, dr \, d\theta\)
    ​\(\displaystyle =\int_{0}^{2\pi}\left[\frac{r^{3}}{3}\right]_{1}^{\sqrt{2}}\,d\theta=\int_{0}^{2\pi} \left(\frac{(\sqrt{2})^{3}}{3}-\frac{1^{3}}{3}\right)\,d\theta=\int_{0}^{2\pi}\left (\frac{2\sqrt{2}}{3}-\frac{1}{3}\right)\,d\theta= 2\pi \left( \frac{2\sqrt{2}}{3} - \frac{1}{3}\right)\)​

  2. Let \(S \subset \mathbb{R}^2\) be the region inside the circle \(x^2 + y^2 = 9\), below the line \(y=x\), above the \(x\)-axis and lying on the right of \(x=1\). Evaluate \(\displaystyle \iint_{S} xy \, dA\).

    We define \(g: D \longrightarrow S\) where \((r,\theta)\mapsto(r\cos\theta,r\sin\theta)\)

    image​

    Now describe \(D\), we know \(0 \leq \theta \leq \frac{\pi}{4}\)

    For each fixed \(\theta\) we enter \(S\) at \(x=1\) and we leave at \(x^2+y^2=9\).
    At \(x=1 \implies r\cos\theta=1 \implies r = \frac{1}{\cos\theta} = \sec\theta\). At \(x^2+y^2=9 \implies r^2=9 \implies r=3\).
    So \(\sec\theta \leq r \leq 3\).

    image​

    Therefore \(\displaystyle \int_{S}xy \, dA = \int_{0}^{\frac{\pi}{4}}\int_{\sec\theta}^{3}(r\cos\theta)( r \sin\theta) r\, dr\, d\theta\)
    ​\(\displaystyle= \int_{0}^{\frac{\pi}{4}}\int_{\sec\theta}^{3} r^{3} \cos\theta \sin\theta \, d r\, d\theta\) \(\displaystyle = \frac{1}{4}\int_{0}^{\frac{\pi}{4}}(81 - \sec^{4}\theta) \cos\theta \sin\theta \, d\theta\)
    ​\(\displaystyle = \frac{1}{4}\int_{0}^{\frac{\pi}{4}}(81\cos\theta \sin\theta - \frac{\sin\theta}{\cos^{3}\theta} ) \, d\theta\) \(\displaystyle= \frac{79}{16}\)

  3. Calculate \(\iint_S (x^2+y^2)^{-3/2} dA\) where S is the planar region bounded below by \(y=1\) and above by \(x^2+y^2=4\).

    image​

    If \(y=1\), \(x^2+1^2=4 \implies x^2=3 \implies x=\pm\sqrt{3}\)
    What are the angles between \((1,0)\) and the vectors \((\sqrt{3},1)\) and \((-\sqrt{3},1)\)?

    Dot product: \((1,0) \cdot (\sqrt{3},1) = \|(1,0)\| \|(\sqrt{3},1)\| \cos\theta\), then \(\cos\theta = \frac{\sqrt{3}}{2} \implies \theta=\frac{\pi}{6}\)

    Same way: \((1,0) \cdot (-\sqrt{3},1) = \|(1,0)\| \|(-\sqrt{3},1)\| \cos\theta\), then \(\cos\theta = -\frac{\sqrt{3}}{2} \implies \theta=\frac{5\pi}{6}\)

    So \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\).

    For each fixed \(\theta\) we enter the region at \(y=1\) and exit at \(x^2+y^2=4\)

    If \(y=1 \leadsto r \sin\theta = 1 \leadsto r = \frac{1}{\sin\theta} = \csc(\theta)\)
    If \(x^2+y^2=4 \leadsto r^2=4 \leadsto r=2\), then \(\csc(\theta) \leq r \leq 2\).

    Thus \(\displaystyle \iint_{S} \frac{1}{(x^{2}+y^{2})^{3/2}}dS = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \int_{\csc\theta}^{2} \frac{1}{r^{3}}\underbrace{rdrd\theta}_{dA}\)
    ​\(\displaystyle = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{\csc\theta}^{2}\frac{1}{r^{2}}drd\theta = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left[-\frac{1}{r}\right]_{\csc\theta}^{2} d\theta= \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(-\frac{1}{2}+ \sin\theta\right )d\theta= \sqrt{3}- \frac{\pi}{3}\)

  4. Let \(f(x,y)=\frac{y^{2}}{\sqrt{x^{2}+y^{2}}}\). Let \(S\) be the planar region lying inside the circle \(x^{2}+y^{2}=2x\), above the X-axis and on the right of \(x=1\). Evaluate \(\displaystyle \iint_{S} f dA\).
    ​\(x^{2}+y^{2}=2x\iff(x-1)^{2}+y^{2}=1\)

    image​

    We have \(0 \le \theta \le \frac{\pi}{4}\), then \((1,0).(1,1) = ||(1,0)||.||(1,1)||\cos\theta\)​\(\implies \cos\theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4}\)

    For a fixed \(\theta\) we enter the region at \(x=1\) and leave the region at \((x-1)^{2}+y^{2}=1\)

    If \(x=1 \implies r\cos\theta=1 \implies r=\sec\theta\)
    If \((x-1)^{2}+y^{2}=1 \implies (r\cos\theta-1)^{2}+r^{2}\sin^{2}\theta=1\)
    ​\(\implies r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta=1\)​\(\implies r^2-2r\cos\theta=0\)​\(\implies r=2\cos\theta\)

    So \(\sec\theta \le r \le 2\cos\theta\)
    Then \(\displaystyle\int_{S} f dA = \int_{0}^{\pi/4}\int_{\sec\theta}^{2\cos\theta}\frac{r^{2}\sin^{2}\theta}{r} . r dr d\theta = \int_{0}^{\pi/4}\int_{\sec\theta}^{2\cos\theta}r^{2}\sin^{2}\theta dr d\theta = \frac{1}{6}\ln(\sqrt{2}+1) - \frac{\sqrt{2}}{90}\)

  5. Let \(S \subset \mathbb{R}^2\) be the parallelogram with vertices \((0,0), (3,1), (4,3), (1,2)\). Evaluate \(\displaystyle\iint_{S} x \, dA\).

    image​

    If we want to integrate with respect to \(x\) and \(y\), we'll need three iterated integrals.
    Instead, let's use a substitution and compute only one integral.
    The idea (since the region is a parallelogram) is to integrate over a square, say \(R = [0,1] \times [0,1]\), using a function \(g: R \to S\) bijective and differentiable.
    And linear (linear transf. transform squares into parallelograms).

    We define \(g: R \to S\) as \(g(1,0)=(3,0)\), \(g(0,1)=(1,2)\)
    ​\(g(u,v)=(x,y) = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = (3u+v, u+2v)\) and \(|D_{g}| = |J_{g}|= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}=5\)

    \(\displaystyle \int_S xdA=\int^1_0\int^1_0(3u+v)\cdot 5\,du\,dv=10\)

  6. Let \(S \subset \mathbb{R}^2\) by the region bounded by the curves \(y=x\), \(xy=1\), \(xy=4\) and \(y=3\). \(y>0\). Evaluate \(\displaystyle\iint_{S} y \, dA\)
    Using the substitution \(u=xy\), \(v=\frac{y}{x}\)

    Determine \(g(u,v)\). We have \(y=vx\). So \(u=x \cdot vx \implies \frac{u}{v}=x^2 \implies x=\sqrt{\frac{u}{v}}\)​\(\implies v = \frac{y}{\sqrt{\frac{u}{v}}} \implies y = \sqrt{u \cdot v}\)
    So \(g(u,v) = \left(\sqrt{\frac{u}{v}}, \sqrt{uv}\right)\) image​

    image​

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