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In the complex domain the theory of integration is quite different. In large part, this is due to Cauchy's integral formula. Long story short, there are many theorems in complex analysis which are not replicated in the real case. So, I would not agree these are the "same" thing. That said, there is a fundamental theorem of calculus for both. As you say, ...


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The innermost integral is with respect to $x$, treating $z$ as a constant. Since this is the case, you might as well go with: $$ x = z\tan\theta \implies \text{d}x = z \sec^2\theta \text{ d}\theta \qquad (*)$$ $$ x = 0 \implies \theta = 0, \ x = z \implies \theta = \dfrac{\pi}{4} $$ $$ \int_{0}^{z} \dfrac{z}{x^2+z^2} \text{ d}x = \int_{0}^{\pi/4} ...


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METHOD 1: We can simplify the problem by noting that $\nabla \times \vec F=0$. Therefore, $\vec F$ is therefore conservative on any connected domain. By Stokes' Theorem, the integral of $\vec F$ over any contour $C$ that bounds a connected domain $S$ is $$\oint_C \vec F\cdot d\vec \ell =\int_S \nabla \times \vec F\cdot \hat ndS$$ Thus, the integral of ...


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The integration path have to be closed in order to use Green's theorem so I assume you mean that $C$ is the closed rectangle $E\to F\to G\to H \to E$. mickep's answer has explained well where the mistake in the calculation. I just want to point out that when in doubt you can check your calculation by performing the line integral. For rectangular paths this ...


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Gravity! Imagine if you have some object at a height $h$. From some physical intuition, you can imagine that, as long as you get back to the same position (and in fact, you could end up at any point with the height $h$!), there will be no energy gain from the (Newtonian) gravity.


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The most common, simple examples of conservative vector fields are the gravitational and electronic fields around a point particle. These are conservative fields, so the concept of "potential energy" of gravitation or static electricity makes sense. This also allows orbits, as @JyrkiLahtonen just pointed out in a commment. (Remember that for a while, ...


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I think there is nothing more to say in such a constraint optimisation problem. However, you can check whether the function actually does what it says it does. For instance, to show that they are maximum points, you convert the function to one-variable equation using the constraint, then differentiate once, and substitute. If the values are -ve, then you're ...


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No. It is correct up to the third line from bottom in your calculation, where you insert the bounds for both $x$ and $y$. You should do that only for $x$. Then integrate the $\cos y$. Also, in your last step, the $\pi$ disappears (but that $\pi$ shouldn't be there if you fix the previous mistake). I get $2e^{-\pi}-2$ as a result.


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I think the following picture can explain your doubt.


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If you know what the Kronecker delta $\delta_{ij}$ and the Levi-Civita symbol $\epsilon_{ijk}$ are, there's an even simpler method than that presented by Kelechi Nze above/below. If this is your first time with these symbols then what I'm going to show you below may look complicated but, believe me, it's a simple and very powerful method, well worth ...


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Here's an (ugly) method: First write $$\boldsymbol{w}=\begin{pmatrix}w_1\\w_2\\w_3 \end{pmatrix}, \boldsymbol{r} = \begin{pmatrix}r_1\\r_2\\r_3 \end{pmatrix}$$ Then compute \begin{equation} (\boldsymbol w \times\boldsymbol r)\cdot (\boldsymbol w \times\boldsymbol r) = w_2^2r_3^3 + w_3^2r_2^2 - 2w_2r_3w_3r_2 \\ + w_1^2r_3^3 + w_3^2r_1^2 - 2w_1r_3w_3r_1 \\ ...


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For part #1: Let's call the two given planes $A$ and $B$, and the desired plane $C$. If $C$ is perpendicular $A$ and $B$, then the normal $N_C$ of $C$ is perpendicular to the normals $N_A$ and $N_B$ of $A$ and $B$. Draw a picture to convince yourself that this is true. So $N_C$ is parallel to the cross product $N_A \times N_B$. In your case, $A$ is the ...


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In the first part you need the cross product of the two plane normals to find the required normal to the plane, so that you can write down the equation In the second part, find the vector joining the two points and do the cross product of this and the normal to the given plane,cso you can write down the equation in a similar way as before.


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Here is what you want to think: Find the line through $(3,0,-1)$ that is orthogonal to each plane (use projection maps). Now you have two lines through a common point, which determines a plane. Think of the three coordinate planes in $\mathbb{R}^{3}$: no two are parallel, but each is perpendicular to both of the others. (this is just part a). For part ...


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If any of $x$, $y$, and $z$ is $0$, then two of them are $0$ and the other is not (so as to satisfy $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$). Thus, $\lambda=0$. Hence, all such solutions are $(x,y,z,\lambda)=(\pm a,0,0,0)$, $(x,y,z,\lambda)=(0,\pm b,0,0)$, and $(x,y,z,\lambda)=(0,0,\pm c,0)$. If none of $x$, $y$, and $z$ is $0$, then ...


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We have: $3(xyz)^2 = 3xy\times yz \times zx = \dfrac{24xyz\times \lambda^3}{a^2b^2c^2}, 3xyz = xyz+xyz+xyz = 2\lambda\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\right)=2\lambda \times 1 = 2\lambda\Rightarrow 2\lambda = 3xyz=\dfrac{24\lambda^3}{a^2b^2c^2}\Rightarrow \lambda^2 = \dfrac{a^2b^2c^2}{12}\Rightarrow \lambda = \dfrac{abc}{2\sqrt{3}}$.


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As you mentioned, the force field $F(x,y,z) = yz\hat{i}+zx\hat{j}+xy\hat{k}$ is a conservative force field with a potential function of $f(x,y,z) = xyz$. Hence, the work done in moving from the origin $O(0,0,0)$ to a point $A(\xi,\eta,\zeta)$ is simply $f(\xi,\eta,\zeta) - f(0,0,0) = \xi\eta\zeta$. Since the point $A(\xi,\eta,\zeta)$ is on the surface of ...


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Hint Rearranging the first display equation motivates reframing the problem as follows: Show that for any $1$-form $\omega$ on $\Bbb R^2 - \{ 0 \}$ there is some constant $\mu$ such that the form $$\eta := \omega - \mu \, d\theta$$ is exact, that is, that there is a function $f \in C^1(\Bbb R^2 - \{ 0 \})$ such that $\eta = df$. On the other hand, ...


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There is a general theory of differentiation for functions between two normed space. However, you may be happy to learn that a function $f \colon \mathbb{R}^n \to \mathbb{R}^m$ is continuously differentiable if and only if each component $f_i \colon \mathbb{R}^n \to \mathbb{R}$ is continuously differentiable, for $i=1,\ldots,m$.


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If you use the Euclidean norm on $\mathbb{R}^n$ then continuity of $Df$ means that for all $\epsilon>0$, there exists a $\delta >0$ such that if $||u - v|| <\delta$ then $||Dfu - Dfv|| < \epsilon$. Edit Using your mapping that you have above, if we let $u = \left [ \begin{array}{c} u_1 \\ u_2 \\ \end{array} \right ]$ and $v = \left [ ...


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$ (x-1)^2 + ( y/3)^2 = 9$. Major , minor axes are a, b Sum of curvatures ( obtained by differentiation) $$ \dfrac{b}{a^2} +\dfrac{a}{b^2},\, a= 1, b=3 $$ Plug in.


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Note: At $(0,0)$, we have that $f(0,0) = 0$. Now, the limit doesn't exists, as along the path $y=x$, for all small enough radius deleted neighborhoods of $(0,0)$, $f(x,y) = 0$ (or, better yet, just look at the path $y=x^2$!), yet along the path $y=x^4$, for all small enough radius deleted neighborhoods of $(0,0)$, $f(x,y) = 1$.


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The proof has to proceed by induction on n because then you are always working with neighborhoods of P0. F1(x1,x2,y1,y2)=0 F2(x1,x2,y1,y2)=0 Have a solution in a neighborhood of P0. (F1->F2) F1(x1,x2,y1,y2)=0 F2(x1,x2,y1,y2,y3)=0 P01 is entirely different than P02. (F1->’F2) See appendix of “Rutherford Aris; Vectors, Tensors, and the Basic Equations of ...


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Change of order of integration we have $$\int_{0}^{1}\int_{\sqrt{x}}^{1}e^{y^3}dydx$$ By changing order of integration, we get area bounded by $x=y^2$, $x=0$, $0\leq y\leq 1$ $$=\int_{0}^{1}\int_{0}^{y^2}e^{y^3}dxdy$$ $$=\int_{0}^{1}\left(\int_{0}^{y^2}e^{y^3}dx\right)dy$$ $$=\int_{0}^{1}e^{y^3}[x]_{0}^{y^2}dy$$ ...


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Let $D$ be an region under cycloid (which are you are finding). So, you should compute an integral: $$\iint_{D} 1 \; dx dy$$ In wikipedia we can find Green's theorem, so if we choose for example $M(x,y)=2x$ and $L(x,y)=y$ we have: $$\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}=2-1=1$$ So: $$\iint_{D}\frac{\partial M}{\partial ...


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NOTE: The solution herein addressed the Originally Posted Question, which questioned the evaluation of the line integral $\oint_C (xy^2\,dx+2x^2\,dy)$. Since this post, the OP edited the question to request verifying evaluation of the line integral $\oint_C (xy^2\,dx+2x^2y\,dy)$. REPLY TO THE ORIGINALLY ASKED QUESTION In using Green's Theorem, ...


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If ${\bf F} = \langle P, Q \rangle$ is conservative, by definition there is some function $f$ such that $${\bf F} = \nabla f = (f_x, f_y).$$ (At least under the modest assumption that $f$ is $C^2$,) by Clairaut's Theorem the mixed partial derivatives of $f$ commute, and so $$P_y = f_{xy} = f_{yx} = Q_x.$$ On the other hand, one can show that the resulting ...


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Adjusting the substitution: $$x=\frac{2}{5} (u-v)$$ $$y=\frac{4u+v}{5}$$ $$\frac{\partial x}{\partial u}=\frac{2}{5} ,\frac{\partial x}{\partial v}=-\frac{2}{5}, \frac{\partial y}{\partial u}=\frac{4}{5},\frac{\partial y}{\partial v}=\frac{1}{5}$$ So the Jacobian is $$\frac{2}{5}\times\frac{1}{5} - (\frac{-2}{5})\times\frac{4}{5}=\frac{10}{25}$$ and not ...


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Hint: $$\int_C f(x,y) dy=\int_a^b f(x(t),y(t)) y'(t) dt$$ Parametrize your line segment and plug it in here and you have your answer.


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It's really simpler evaluate it in polar coordinates. $ds=3\,d\phi$, and integral is $$ \int_0^{2\pi} 9\sin^2\phi\cdot 3\,d\phi = 27\pi $$


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This vector field is conservative. you just check that the derivative wrt y of the first component is the same as the derivative wrt x of the second component. The primitive function of this vector field is xe^(xy)+c. Thus the integral of this over the upper semicircle is the value of the primitive at (1,0) minus the value at (-1,0). It is 2.


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Writing $\int_\phi {\bf F}$ is a notation: you are integrating along the path. By definition this is $\int_a^b {\bf F}(\phi(t))\cdot \phi'(t)\,{\rm d}t$. By definition of the dot product in $\Bbb R^n$ this is $\int_a^b \sum F_i(\phi(t))\phi_i'(t)\, {\rm d}t$. Since integrals are additive, this reads $\sum \int_a^b F_i(\phi(t))\phi_i'(t)\, {\rm d}t$. Now we ...


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How about $$\iint_{A} dxdy= \text{Area of A}=\frac{\pi}{6}$$ If you must use Cartesians (although I stress that the above is correct!) then at least by symmetry we can take double the integral of the sector with $x>0$. It will be easier to integrate over $y$ first, since we won't need to change our boundaries. We have $\sqrt{3}x<y<\sqrt{1-x^{2}}$, ...


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One parametrization of $\mathscr{C}$ is $$ \begin{bmatrix}x\vphantom{\frac1{\sqrt2}}\\y\vphantom{\frac1{\sqrt2}}\\z\vphantom{\frac1{\sqrt2}}\end{bmatrix} =a\cos(\theta)\begin{bmatrix}\frac1{\sqrt2}\\\lower{2pt}{0}\vphantom{\frac1{\sqrt2}}\\-\frac1{\sqrt2}\end{bmatrix} ...


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When you're using Green's theorem to get the above formula for the area enclosed by the curve, you have to assume that the curve is oriented counterclockwise. I guess this is why the absolute value comes in, because the given parametrization doesn't specify the orientation. Consider for example the case of the unit circle where $f(\phi)=\sin\phi$ and ...


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For "well-behaved" functions $f$ in a well-behaved region such as ours, the only candidates for maximumhood/minimumhood are (i) the point(s) in the interior where $f_x=0$ and $f_y=0$ and (ii) points on the boundary. In our case, the only interior point, indeed the only point, where the partials both vanish is $(0,0)$. That's one candidate. You report that ...


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Since the plane passes through the origin, the curve of intersection is a circle of radius $a$ and so we may take our surface to be a disk of radius $a$. (Note that Stokes' Theorem allows you to take $\textit {any}$ surface whose boundary is the curve over which the line integral is to be taken. We choose the disk because it is the easiest). Now, observe ...


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You are correct that $S$ is not a smooth surface in $\mathbb{R}^3$. Here is a rough sketch of a proof. Lemma 1: $S$ is a surface iff $T = \{(x,y,z)\in \mathbb{R}^3: x^3 + y^3 + z^3 =0 \}$ is a surface. Proof idea: Consider $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ with $f(x,y,z) = (x, \sqrt[3]{2}y, \sqrt[3]{3}z)$. Then $f$ is a diffeomorphism, and ...


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Hint: Let $\gamma: [a,b] \to D(f)$ and $\sigma:[c,d] \to D(f)$ be two distinct parametrization of $C$ with the same orientation. Hence there exist a continuous function $\varphi: [c,d] \to [a,b]$, strictly increasing with $\varphi(c)=a, \ \varphi(d)=b$, such that $\sigma = \gamma \circ \varphi$. Now use the definition of $\int_C f$ and the fact that ...


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I suggest taking $s = \frac{\xi}{b}$ so the denominator looks as the denominator of the integral you know how to solve. Then, think using the condition "for any $x \in \mathbb{R} $".


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First of all, write formula from $\textbf{[1]}$ for $x=yb$, that means: $$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; d\xi =\pi e^{-b|y|}$$ Next change of variables in integral: $$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; d\xi$$ Putting $\xi=\frac{\xi}{b}$: $$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; ...


1

No, there is no point trying to render this into Cartesian form: it is better to stick to parametric form Hint...if you can find the right $t$ values for the limits, you only need $$\int y\frac{dx}{dt}dt$$


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div$\vec F=\bigg(\frac{\partial(4x)}{\partial x}+\frac{\partial (-y^2)}{\partial y}+\frac{\partial (z^2)}{\partial z}\bigg)=\bigg(4-2y+2z\bigg)$ $$\iint_{S}\vec F\cdot \hat n \;ds=\iint_{V}div\vec F \;dv=\iiint_{V}\bigg(4-2y+2z\bigg) \;dxdydz$$ $$x=\rho \cos \varphi,\; y=\rho \sin \varphi, \; z=z$$ $$0\leq z \leq3, \; 0\leq \rho \leq 2, \; ...


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You must use the Gauss Theorem (divergent theorem), since it deals with flux: $$ \nabla F = 4-4y+2z $$ $$ \int_{V} \nabla F dV = \int\int \vec{F}\vec{n}dS $$ Your volume is define by the cylinder $x^2 + y^2 = 4$ and $0 \leq z \leq 3$.


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Recall that, for a differentiable function, the directional derivative of $f$ in the (unit) direction $u$, at a point $(x, y, z)$, is given by $$D_u f(x, y, z) = \nabla f(x, y, z) \cdot u.$$ Also recall the following identity: $$v \cdot w = |v||w|\cos(\theta),$$ where $\theta$ is the angle between $v$ and $w$. Since $|u| = 1$, it follows that $$D_u f(x, y, ...


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while converting in polar co-ordinate you have directly taken dxdx=drd(theta), you have to use polar transformation Jacobin to do so, that is dxdx=rdr*d(theta), then integrate normally answer will be (ip/2).


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Switching to polar coordinates, the Jacobian is given by $ |J|$ where $$ J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & \sin\theta \\ ...


1

Multiply Jacobian in integrand $$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^3\bigg)dr\bigg]d\theta$$


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The rate of increase of the function $f(\vec{x})$ in the direction of $\vec{u}$, a unit vector, is $$D :=\lim_{t \to 0} \frac{f(\vec{x} + t\vec{u}) - f(\vec{x})}{t}$$ Now let $F(t) = f(\vec{a} + t\vec{u})$, then $$D = \frac{dF}{dt}(0) = \frac{\partial f}{\partial x_1}(\vec{a}) \, \frac{dx_1}{dt} + ... + \frac{\partial f}{\partial x_n}(\vec{a}) \, ...


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I'm not sure if you're asking for mathematical reasoning or a physical motivation for surface integrals. If it's the latter, you'll find plenty motivation for surface integrals when dealing with flux. Let's start with a simpler example. Assume you have some metal plate that has a temperature gradient on it. You know the temperature $T(x,y)$ at each point, ...



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