Line Integrals

Assume that \(\varepsilon > 0\) is given. Let \(\delta_1 > 0\) be a real number such that \[\left|\int_{\gamma} f\,ds-S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1}\right|\leq\frac{\varepsilon}4\quad\quad\quad\quad\quad (1)\] whenever \(\delta(t_0,\dots, t_{n-1}) < \delta_1\).

Recall that the line segment between the points \(\gamma_{i-1}\) and \(\gamma_i\) has the length \(l_i=\sqrt{\left(\phi(t_i)-\phi(t_{i-1})\right)^2+\left(\phi(t_i)-\psi(t_{i-1})\right)^2}\). According to the mean-value theorem we know that there exists a number \(\xi_{i-1}\in [t_{i-1},t_i]\) such that \(\phi(t_i)-\phi(t_{i-1})=\phi^{\prime}(\xi_{i-1})\cdot (t_i-t_{i-1})\). Similarly, there exists \(\eta_{i-1}\in [t_{i-1},t_i]\) such that \(\psi(t_i)-\psi(t_{i-1})=\psi^{\prime}(\eta_{i-1})\cdot (t_i-t_{i-1})\). Because of the continuity of \(f\), \(\phi^{\prime}\) and \(\psi^{\prime}\) on \([a,b]\), there exists a real number \(M > 1\) such that \(\max_{[a,b]}|f|\leq M\), \(\max_{[a,b]}|\phi^{\prime}|\leq M\), and \(\max_{[a,b]}|\psi^{\prime}|\leq M\). Also, there exists \(\delta_2 > 0\) such that \[\left|\phi^{\prime}(\xi_{i-1})-\phi^{\prime}(c_{i-1})\right|\leq \frac{\varepsilon^2}{16M^3(b-a)^2},\quad\quad\quad\mbox{ and }\quad\quad\quad \left|\psi^{\prime}(\eta_{i-1})-\psi^{\prime}(c_{i-1})\right|\leq \frac{\varepsilon^2}{16M^3(b-a)^2}\] whenever \(\delta(t_0,\dots, t_n) < \delta_2\).

Let us take \(\delta=\min\{\delta_1,\delta_2\}\) and assume that \(t_0,\dots, t_n\) is any partition with \(\delta(t_0,\dots, t_n) < \delta\). If \(c_0\), \(\dots\), \(c_{n-1}\) is any choice of points such that \(c_i\in[t_i,t_{i+1}]\) we have that: \(|S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1})-\int_{\gamma}f\,ds|\leq \varepsilon/4\). We also have that \begin{eqnarray*}S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1})&=&\sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot l_i\newline &=& \sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot \sqrt{\phi’(c_i)^2+\psi’(c_i)^2}\cdot (t_i-t_{i-1})\\ &&+\sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot \left(l_i-\sqrt{\phi’(c_i)^2+\psi’(c_i)^2}\cdot (t_i-t_{i-1})\right)\\ &=& \sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot \sqrt{\phi’(c_i)^2+\psi’(c_i)^2}\cdot (t_i-t_{i-1})\\&&+ \sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot \left(\sqrt{\phi’(\xi_{i-1})^2+\psi’(\eta_{i-1})^2}-\sqrt{\phi’(c_i)^2+\psi’(c_i)^2} \right)\cdot (t_i-t_{i-1}). \quad\quad\quad (2) \end{eqnarray*} The first sum on the right-hand side is the integral sum for \(\int_a^b f(\phi(t),\psi(t))\sqrt{\phi^{\prime}(t)^2+\psi^{\prime}(t)^2}\,dt\). Therefore there exists \(\delta_3 > 0\) such that \(\delta(t_0,\dots, t_n) < \delta_3\) implies \[\left|\sum_{i=0}^{n-1} f(\phi(c_i),\psi(c_i))\cdot \sqrt{\phi’(c_i)^2+\psi’(c_i)^2}\cdot (t_i-t_{i-1})-\int_a^b f(\phi(t),\psi(t))\sqrt{\phi^{\prime}(t)^2+\psi^{\prime}(t)^2}\,dt \right| < \frac{\varepsilon}4.\] We will prove that the second sum in (2) is negligible. First, we notice that for \(A,B\geq 0\) the following inequality holds: \(|\sqrt A-\sqrt B|\leq \sqrt{|A-B|}\), hence \begin{eqnarray*}\sqrt{\phi’(\xi_{i-1})^2+\psi’(\eta_{i-1})^2}-\sqrt{\phi’(c_i)^2+\psi’(c_i)^2}&\leq& \left(\left|\phi^{\prime}(\xi_{i-1})-\phi^{\prime}(c_{i-1})\right|\cdot\left(|\phi^{\prime}(\xi_{i-1})|+|\phi^{\prime}(c_{i-1})|\right) \right.\\ &&\left .+ \left|\psi^{\prime}(\eta_{i-1})-\psi^{\prime}(c_{i-1})\right|\cdot\left(|\psi^{\prime}(\eta_{i-1})|+|\psi^{\prime}(c_{i-1})|\right) \right)^{\frac12} \\ &\leq& \sqrt{\frac{\varepsilon^2}{16M^3(b-a)^2}\cdot (2M+2M)}\leq \frac{\varepsilon}{2M(b-a)}.\end{eqnarray*} Therefore \begin{eqnarray*} &\sum_{i=0}^{n-1}& f(\phi(c_i),\psi(c_i))\cdot \left(\sqrt{\phi’(\xi_{i-1})^2+\psi’(\eta_{i-1})^2}-\sqrt{\phi’(c_i)^2+\psi’(c_i)^2} \right)\cdot (t_i-t_{i-1})\\&\leq& \frac{\varepsilon}{2M(b-a)}\cdot \sum_{i=0}^{n-1}\left|f(\phi(c_i),\psi(c_i))\right|\cdot (t_i-t_{i-1})\leq \frac{\varepsilon}{2M(b-a)}\cdot \sum_{i=0}^{n-1}M\cdot (t_i-t_{i-1})\\ &=& \frac{\varepsilon}2. \end{eqnarray*} Therefore:\begin{eqnarray*} \left|S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1})-\int_a^b f(\phi(t),\psi(t))\sqrt{\phi^{\prime}(t)^2+\psi^{\prime}(t)^2}\,dt\right|\leq \frac{\varepsilon}2+\frac{\varepsilon}4=\frac{3\varepsilon}4. \end{eqnarray*} Together with (1) this gives: \begin{eqnarray*} \left|\int_{\gamma} f\,ds-\int_a^b f(\phi(t),\psi(t))\sqrt{\phi^{\prime}(t)^2+\psi^{\prime}(t)^2}\,dt\right|&\leq &\left|\int_{\gamma} f\,ds-S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1})\right|\\&&+\left|S(f,t_0,\dots, t_n,c_0,\dots, c_{n-1})-\int_a^b f(\phi(t),\psi(t))\sqrt{\phi^{\prime}(t)^2+\psi^{\prime}(t)^2}\,dt\right|\leq \varepsilon. \end{eqnarray*} Since this holds for every \(\varepsilon > 0\) the required equality holds.

Let us first parametrize the curve \(\gamma\). We do this in the following way: \begin{eqnarray*} x&=&1+2\cos\theta\\ y&=&2\sin \theta\\ 0\leq &\theta&\leq \pi. \end{eqnarray*} The integral now becomes \begin{eqnarray*} \int_{\gamma}f\,ds&=&\int_0^{\pi} (1+2\cos\theta)^2\cdot 2\sin\theta\cdot\sqrt{x’(\theta)^2+y’(\theta)^2}\,d\theta= \int_0^{\pi}(1+2\cos\theta)^2\cdot2\sin\theta\cdot\sqrt{(-2\sin\theta)^2+(2\cos\theta)^2}\,d\theta\\ &=&4\int_0^{\pi}\left(1+4\cos\theta+4\cos^2\theta\right)\sin\theta\,d\theta. \end{eqnarray*} Let us use the substitution \(\cos\theta=u\) in the last integral. Then the variable \(u\) ranges from \(1\) to \(-1\), and \(\sin\theta\,d\theta=-du\). Hence \[ \int_{\gamma}f\,ds=4\int_{-1}^1\left(1+4u+4u^2\right)\,du=4\cdot \left.\left(u+u^2+\frac{4u^3}3\right)\right|_{-1}^1=4\cdot \left(2+\frac{8}3\right)=\frac{56}3. \]

Let \(\varepsilon > 0\) be given. Since \(\overrightarrow F\) is integrable with respect to \(\gamma\) there exists \(\delta_1 > 0\) such that \[\left|S(\overrightarrow{F},t_0,\dots, t_n,c_0,\dots, c_{n-1})-\int_{\gamma}\overrightarrow F\cdot d\overrightarrow r\right| < \frac{\varepsilon}2\quad\quad\quad\quad\quad (1)\] whenever \(\delta(t_0,\dots, t_n)\leq \delta_1\). For any two values \(\tau_1,\tau_2\in[a,b]\) we have \(\overrightarrow r(\tau_2)-\overrightarrow r(\tau_1)=\langle \phi(\tau_2)-\phi(\tau_1),\psi(\tau_2)-\psi(\tau_1),\theta(\tau_2)-\theta(\tau_1)\rangle\). According to the mean-value theorem, there are numbers \(\xi\), \(\eta\), \(\zeta\in [\tau_1,\tau_2]\) such that \(\phi(\tau_2)-\phi(\tau_1)=\phi^{\prime}(\xi)\cdot (\tau_2-\tau_1)\), \(\psi(\tau_2)-\psi(\tau_1)=\psi^{\prime}(\eta)\cdot (\tau_2-\tau_1)\), and \(\theta(\tau_2)-\theta(\tau_1)=\theta^{\prime}(\zeta)\cdot (\tau_2-\tau_1)\). Because of continuity of \(\phi\), \(\psi\), \(\theta\), and \(\overrightarrow F\) there is a real number \(M > 0\) such that \[\max_{[a,b]}\left\{\left|\phi(t)\right|,\left|\psi(t)\right|,\left|\theta(t)\right|,\left|\overrightarrow F\left(\overrightarrow r(t)\right)\right|\right\} < M,\;\;\;\mbox{and }\delta_2 > 0 \mbox{ such that }\] \[ \max_{|\mu_1-\mu_2| < \delta_2}\left\{\left|\phi^{\prime}(\mu_1)-\phi^{\prime}(\nu_1)\right|, \left|\psi^{\prime}(\mu_1)-\psi^{\prime}(\nu_1)\right|, \left|\theta^{\prime}(\mu_1)-\theta^{\prime}(\nu_1)\right|\right\}\leq \frac{\varepsilon}{4\sqrt 3M(b-a)}. \] Let us now study the integral sum for the line integral \(\int_{\gamma} \overrightarrow F\cdot d\overrightarrow r\). We will use that \[\overrightarrow r(t_{i+1})-\overrightarrow r(t_i)=\langle \phi(t_{i+1})-\phi(t_i),\psi(t_{i+1})-\psi(t_i),\theta(t_{i+1})-\theta(t_i)\rangle=(t_{i+1}-t_i) \langle \phi^{\prime}(\xi_i),\psi^{\prime}(\eta_i),\theta^{\prime}(\zeta_i)\rangle,\] for some numbers \(\xi_i,\eta_i,\zeta_i\in[t_i,t_{i+1}]\). This implies that \begin{eqnarray*} S(\overrightarrow{F},t_0,\dots, t_n,c_0,\dots, c_{n-1})&=&\sum_{i=0}^{n-1} \overrightarrow F(P_i)\cdot \left(\overrightarrow r(t_{i+1})-\overrightarrow r(t_i)\right)=\sum_{i=0}^{n-1}\overrightarrow F(P_i)\cdot \langle \phi’(\xi_i),\psi’(\eta_i),\theta’(\zeta_i)\rangle (t_{i+1}-t_i)\\ &=&\sum_{i=0}^{n-1} \overrightarrow F(\phi(c_i),\psi(c_i),\theta(c_i))\cdot \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle (t_{i+1}-t_i)\\&&+\sum_{i=0}^{n-1} \overrightarrow F(P_i)\cdot \left(\langle \phi’(\xi_i),\psi’(\eta_i),\theta’(\zeta_i)\rangle - \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle \right)(t_{i+1}-t_i). \quad\quad\quad (2) \end{eqnarray*} The first sum on the right-hand side of (2) is the integral sum for the definite integral \(\int_a^b \overrightarrow F(\overrightarrow r(t))\cdot \overrightarrow{r^{\prime}}(t)\,dt\), hence there exists \(\delta_3 > 0\) such that \[\left|\int_a^b \overrightarrow F(\overrightarrow r(t))\cdot\overrightarrow{r^{\prime}}(t)\,dt-\sum_{i=0}^{n-1} \overrightarrow F(\phi(c_i),\psi(c_i),\theta(c_i))\cdot \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle (t_{i+1}-t_i) \right| < \frac{\varepsilon}4\] whenever \(\delta(t_0,\dots, t_n)\leq \delta_3\).

Let us now study the second sum on the right-hand side of (2). We have that \(\left|\overrightarrow F(P_i)\right|\leq M\) and \begin{eqnarray*}&&\left|\langle \phi’(\xi_i),\psi’(\eta_i),\theta’(\zeta_i)\rangle - \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle\right| \\ &=& \sqrt{\left(\phi’(\xi_i)-\phi’(c_i)\right)^2+ \left(\psi’(\eta_i)-\psi’(c_i)\right)^2+\left(\theta’(\zeta_i)-\theta’(c_i)\right)^2}\leq \frac{\varepsilon}{4M(b-a)}. \end{eqnarray*} Therefore: \begin{eqnarray*}&& \left|\sum_{i=0}^{n-1} \overrightarrow F(P_i)\cdot \left(\langle \phi’(\xi_i),\psi’(\eta_i),\theta’(\zeta_i)\rangle - \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle \right)(t_{i+1}-t_i) \right|\\ &\leq& \sum_{i=0}^{n-1}\left| \overrightarrow F(P_i)\cdot \left(\langle \phi’(\xi_i),\psi’(\eta_i),\theta’(\zeta_i)\rangle - \langle \phi’(c_i),\psi’(c_i),\theta’(c_i)\rangle \right)(t_{i+1}-t_i)\right|\\ &\leq& \sum_{i=0}^{n-1}M\cdot \frac{\varepsilon}{4M(b-a)}\cdot (t_{i+1}-t_i)=\frac{\varepsilon}4. \end{eqnarray*} The inequality (2) now implies: \[\begin{eqnarray*}\end{eqnarray*}\begin{eqnarray*}&& \left|S(\overrightarrow{F},t_0,\dots, t_n,c_0,\dots, c_{n-1})-\int_a^b \overrightarrow F\left(\phi(t),\psi(t),\theta(t)\right)\cdot \overrightarrow{r^{\prime}}(t)\,dt \right| &\leq&\frac{\varepsilon}4+\frac{\varepsilon}4=\frac{\varepsilon}2. \end{eqnarray*}\] This together with inequality (1) implies: \begin{eqnarray*} \left|\int_{\gamma}\overrightarrow{F}\cdot d\overrightarrow r-\int_a^b \overrightarrow F\left(\phi(t),\psi(t),\theta(t)\right)\cdot \overrightarrow{r^{\prime}}(t)\,dt\right|\leq\varepsilon. \end{eqnarray*} Since this holds for every \(\varepsilon > 0\) the proof of the theorem is complete.

We will use the following parametrization of the circle: \begin{eqnarray*} x&=&3\cos\theta\\ y&=&3\sin\theta\\ 0\leq&\theta&\leq 2\pi. \end{eqnarray*} Then we have \begin{eqnarray*} \int_{\gamma} \overrightarrow F\cdot d\overrightarrow r&=&\int_0^{2\pi} \langle 3\sin\theta, 6\cos\theta\rangle\cdot \langle -3\sin\theta,3\cos\theta\rangle\,d\theta=\int_0^{2\pi} \left(-9\sin^2\theta+18\cos^2\theta\right)\,d\theta\\ &=&\int_0^{2\pi}\frac{-9\cdot (1-\cos(2\theta))+18(1+\cos(2\theta))}2\,d\theta=\frac 92+\frac{27}2\int_0^{2\pi}\cos(2\theta)\,d\theta=\frac92. \end{eqnarray*}

According to the chain rule we have that the function \[\phi(t)=f\left(\overrightarrow r(t)\right)\] satisfies: \[\phi^{\prime}(t)=\nabla f(\overrightarrow r(t))\cdot \overrightarrow{r^{\prime}}(t)=\overrightarrow F(\overrightarrow r(t))\cdot \overrightarrow{r^{\prime}}(t),\] hence the fundamental theorem of calculus implies: \[f(\overrightarrow r(b))-f(\overrightarrow r(a))=\phi(b)-\phi(a)=\int_a^b \phi^{\prime}(t)\,dt=\int_a^b \nabla f(\overrightarrow r(t))\cdot\overrightarrow{r^{\prime}}(t)\,dt=\int_{\gamma}\nabla f\cdot d\overrightarrow r.\]

We start by writing the integral as the sum of two integrals \[\int_C\langle xy,y^2, x+y+z\rangle\cdot \,d\overrightarrow r=\int_{C_1} \langle xy,y^2, x+y+z\rangle\cdot \,d\overrightarrow r+\int_{C_2}\langle xy,y^2, x+y+z\rangle\cdot \,d\overrightarrow r,\] where \(C_1\) is the arc between \((1,0,0)\) and \((0,1,0)\) and \(C_2\) is the arc between \((0,1,0)\) and \((0,0,1)\).

We will first evaluate the integral over \(C_1\). The parametrization of the curve \(C_1\) is: \begin{eqnarray*} x&=&\cos \theta\\ y&=&\sin\theta\\ z&=&0\\ 0\leq&\theta&\leq \frac{\pi}2. \end{eqnarray*} We now have \(d\overrightarrow r=\langle -\sin\theta,\cos\theta,0\rangle\,d\theta\) and the integral becomes: \begin{eqnarray*}\int_{C_1}\langle xy,y^2, x+y+z\rangle\cdot \,d\overrightarrow r&=&\int_0^{\frac{\pi}2}\langle \sin\theta\cos\theta,\sin^2\theta,\sin\theta+\cos\theta\rangle\cdot \langle -\sin\theta,\cos\theta,0\rangle\,d\theta\\&=& \int_0^{\frac{\pi}2}\left(-\sin^2\theta\cos\theta+\sin^2\theta\cos\theta\right)\,d\theta=0. \end{eqnarray*}

A parametrization of the curve \(C_2\) is: \begin{eqnarray*} x&=&0\\ y&=&\cos\theta\\ z&=&\sin\theta\\ 0\leq&\theta&\leq\frac{\pi}2. \end{eqnarray*} For this parametrization we have \(d\overrightarrow r=\langle 0,-\sin\theta,\cos\theta\rangle\,d\theta\) and the integral is: \begin{eqnarray*} \int_{C_2}\langle xy,y^2,x+y+z\rangle\cdot\,d\overrightarrow r &=& \int_0^{\frac{\pi}2}\langle 0,\cos^2\theta,\cos^2\theta+\sin\theta\rangle\cdot\langle 0,-\sin\theta,\cos\theta\rangle\,d\theta\\&=& \int_0^{\frac{\pi}2}-\cos^2\theta\sin\theta\,d\theta+\int_0^{\frac{\pi}2}\left(1-\sin^2\theta+\sin\theta\right)\cos\theta\,d\theta. \end{eqnarray*} The first integral on the right-hand side can be solved using the substitution \(\cos\theta=u\) and the second one using the substitution \(\sin\theta=v\). In the first integral the new variable \(u\) ranges from \(1\) to \(0\). We also have \(-\sin\theta\,d\theta=du\). Therefore \[\int_0^{\frac{\pi}2}-\cos^2\theta\sin\theta\,d\theta=-\int_0^1u^2\,du=-\frac13.\] Similarly, \[\int_0^{\frac{\pi}2}\left(1-\sin^2\theta+\sin\theta\right)\cos\theta\,d\theta=\int_0^1\left(1-v^2+v\right)\,dv=\frac76.\] Thus \(\int_{C_2}\langle xy,y^2,x+y+z\rangle\cdot\,d\overrightarrow r=\frac76-\frac13=\frac56\) and \[\int_C\langle xy,y^2,x+y+z\rangle\cdot\,d\overrightarrow r=\frac56.\]

We first want to parametrize the curve. The curve consists of all those points \((x,y,z)\) that satisfy both equations. Since both \(z=x^2+4y^2\) and \(z=6-6x-4y\) must be satisfied, we must have \(6-6x-4y=x^2+4y^2\) which is equivalent to \[(x+3)^2+(2y+1)^2=1.\quad\quad\quad\quad\quad (1)\] This means that all points \((x,y,z)\) on the described curve must satisfy (1). The curve can be parametrized in the following way: \begin{eqnarray*} x&=&4\cos\theta-3\\ y&=&2\sin\theta-\frac12\\ z&=&26-24\cos\theta-8\sin\theta\\ 0\leq &\theta&\leq 2\pi. \end{eqnarray*} The following equality holds for this parametrization: \[d\overrightarrow r=\langle -4\sin\theta, 2\cos\theta, 24\sin\theta-8\cos\theta\rangle\,d\theta.\]

The integral now becomes: \begin{eqnarray*} \int_{\gamma}\overrightarrow F\cdot d\overrightarrow r&= & \int_0^{2\pi}\left(2\sin\theta-\frac12\right)\cdot\left(20\sin\theta-6\cos\theta\right)\,d\theta=40\pi. \end{eqnarray*}

We start by writing the integral as the sum of two integrals \[\int_C\left(x^2+y\right)\,ds=\int_{C_1} \left(x^2+y\right)\,ds+\int_{C_2}\left(x^2+y\right)\,ds,\] where \(C_1\) is the line segment between \((0,0)\) and \((1,1)\), and \(C_2\) is the portion of the circle between \((1,1)\) and \((2,0)\).

We will first evaluate the integral over the line segment. The parametrization of the line segment is \(x=t\), \(y=t\), \(0\leq t\leq 1\). Then we have \(ds=\sqrt{x^{\prime}(t)^2+y^{\prime}(t)^2}\,dt=\sqrt2\,dt\). The integral becomes \[\int_{C_1}\left(x^2+y\right)\,ds=\int_0^1\left(t^2+t\right)\cdot \sqrt 2\,dt=\sqrt2\cdot \left(\frac13+\frac 12\right)=\frac{5\sqrt 2}6.\]

Let us now parametrize the curve \(C_2\). We can do this in the following way: \begin{eqnarray*} x&=&\sqrt2\sin \theta\\ y&=&\sqrt2\cos\theta\\ \frac{\pi}4\leq&\theta&\leq \frac{\pi}2. \end{eqnarray*} We now have \(ds=\sqrt{x^{\prime}(t)^2+y^{\prime}(t)^2}\,dt=\sqrt{2}\,d\theta\). The integral becomes: \begin{eqnarray*}\int_{C_2}\left(x^2+y\right)\,ds&=&\sqrt2\int_{\frac{\pi}4}^{\frac{\pi}2}\left(\sin^2\theta+\cos\theta\right)\,d\theta=\sqrt 2\int_{\pi/4}^{\pi/2}\frac{1-\cos(2\theta)}2\,d\theta+\sqrt 2\int_{\pi/4}^{\pi/2}\cos\theta\,d\theta\\&=&\frac{\pi\sqrt2}8-\frac{\sqrt2}4\left(\sin\pi-\sin\frac{\pi}2\right)+\sqrt2\cdot\left(\sin\frac{\pi}2-\sin\frac{\pi}4\right)\\ &=& \frac{\pi\sqrt 2}8+\frac{\sqrt 2}4+\sqrt 2-1=\frac{\pi\sqrt 2}8+\frac{5\sqrt 2}{4}-1. \end{eqnarray*} Thus \[\int_C\left(x^2+y\right)\,ds=\frac{5\sqrt 2}6+\frac{\pi\sqrt 2}8+\frac{5\sqrt 2}{4}-1= \frac{\pi\sqrt 2}8+\frac{25\sqrt 2}{12}-1.\]