# Tag Info

7

Integrating by parts twice, we get \begin{align} &\int_0^L\sin(x)\,e^{-ax}\,\mathrm{d}x\tag{0}\\ &=-\int_0^L e^{-ax}\,\mathrm{d}\cos(x)\tag{1}\\ &=1-e^{-aL}\cos(L)-a\int_0^L\cos(x)\,e^{-ax}\,\mathrm{d}x\tag{2}\\ &=1-e^{-aL}\cos(L)-a\int_0^L e^{-ax}\,\mathrm{d}\sin(x)\tag{3}\\ ... 5 according to the shape of the area of integration and the shape of the function that is under integral, the easiest answer is to define variables u=x+y and v=x-ythen we have:\frac{1}{J}=\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}\Rightarrow |J|=\frac{1}{2}$$and the borders of the area of integration are ... 4 You are correct that S is not a smooth surface in \mathbb{R}^3. Here is a rough sketch of a proof. Lemma 1: S is a surface iff T = \{(x,y,z)\in \mathbb{R}^3: x^3 + y^3 + z^3 =0 \} is a surface. Proof idea: Consider f:\mathbb{R}^3\rightarrow \mathbb{R}^3 with f(x,y,z) = (x, \sqrt[3]{2}y, \sqrt[3]{3}z). Then f is a diffeomorphism, and ... 3 Hint: Let \gamma: [a,b] \to D(f) and \sigma:[c,d] \to D(f) be two distinct parametrization of C with the same orientation. Hence there exist a continuous function \varphi: [c,d] \to [a,b], strictly increasing with \varphi(c)=a, \ \varphi(d)=b, such that \sigma = \gamma \circ \varphi. Now use the definition of \int_C f and the fact that ... 3 If you know what the Kronecker delta \delta_{ij} and the Levi-Civita symbol \epsilon_{ijk} are, there's an even simpler method than that presented by Kelechi Nze above/below. If this is your first time with these symbols then what I'm going to show you below may look complicated but, believe me, it's a simple and very powerful method, well worth ... 3 NOTE: The solution herein addressed the Originally Posted Question, which questioned the evaluation of the line integral \oint_C (xy^2\,dx+2x^2\,dy). Since this post, the OP edited the question to request verifying evaluation of the line integral \oint_C (xy^2\,dx+2x^2y\,dy). REPLY TO THE ORIGINALLY ASKED QUESTION In using Green's Theorem, ... 3 Recall that, for a differentiable function, the directional derivative of f in the (unit) direction u, at a point (x, y, z), is given by$$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, ...

3

Hint: I think that Stokes' theorem is overkill here. We can get a simple parametrization: $${\bf r}(t) = (\cos t, \sin t, -1+\cos t), \quad 0 \leq t< 2\pi,$$and all the functions involved are simple too. Just write ${\rm d}x = -\sin t\,{\rm d}t$, etc, substitute and compute. Use and abuse of periodicity. A lot of terms should cancel out, it seems.

2

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 $x \in \mathbb{R}$".

2

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 $\xi=\frac{\xi}{b}$: $$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; ... 2 I'm not sure if you're asking for mathematical reasoning or a physical motivation for surface integrals. If it's the latter, you'll find plenty motivation for surface integrals when dealing with flux. Let's start with a simpler example. Assume you have some metal plate that has a temperature gradient on it. You know the temperature T(x,y) at each point, ... 2 You can consider that area as the sum of two integrals:$$\int_{C}r ~ dr ~d\theta = \int_{C_1}r ~ dr ~d\theta + \int_{C_2} r ~ dr ~d\theta = 2 \int_{C_1} r ~dr ~d\theta$$Where C_1 is the part of the curve which has a positive x coordinate. For this part of the curve \theta varies in the interval [-\frac{\pi}{4},\frac{\pi}{4}]. So the integral that ... 2 Hint Rearranging the first display equation motivates reframing the problem as follows: Show that for any 1-form \omega on \Bbb R^2 - \{ 0 \} there is some constant \mu such that the form$$\eta := \omega - \mu \, d\theta$$is exact, that is, that there is a function f \in C^1(\Bbb R^2 - \{ 0 \}) such that \eta = df. On the other hand, ... 2$$\int_{-3}^{1}\int_{-\sqrt{x+3}}^{\sqrt{x+3}}dydx + \int_{1}^{5} \int_{-\sqrt{5-x}}^{\sqrt{5-x}}dydx=\\2 (\int_{-3}^{1}\int_{0}^{\sqrt{x+3}}dydx + \int_{1}^{5} \int_{0}^{\sqrt{5-x}}dydx)$$2 This really isn't a Jacobian problem. You don't need to change your variables of integration, except for u-substitutions. Observe that \begin{eqnarray*} \int_0^1 y \left (\int_0^{y^2} \cos (x-y^2) dx\right ) dy & = & \int_0^1 y \left ( \sin (x- y^2) |_0^{y^2} \right ) dy \\ & = & \int_0^1 y(\sin 0 - \sin(-y^2)) dy \\ & = & ... 2 So, you know it is a torus: then, your parametrization should be$$ \begin{cases} x = \big(\, 2 + r \cos \theta \, \big) \cos \phi \\ y = \big(\, 2 + r \cos \theta \, \big) \sin \phi \\ z = r \sin \theta \end{cases} $$where r \in [0,1] and \theta, \phi \in [0, 2 \pi), and with jacobian$$ \big|\,J\,\big| = \left| \begin{vmatrix} \frac{\partial ...

2

I guess you need to evaluate the integral $$\iint_A1\,dxdy,$$ where $A=\psi(\{(r, \theta)\in[0, \infty)\times[0,2\pi]:\pi/4\leq\theta\leq3/4\,\pi \quad\text{and}\quad0\leq r\leq2\})$ and $\psi$ is the polar coordinates map. Then $A=\{(x,y)\in \Bbb R^2: -\sqrt2\leq x\leq\sqrt2\quad \text{and}\quad |x|\leq y\leq\sqrt{4-x^2}\}$. So by noticing that $A$ is a ...

2

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 (\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 \\ ...

2

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: $$\iint_{D}\frac{\partial M}{\partial ... 2 Hint:$$(x+y)^9(x-y)^9=((x+y)(x-y))^9=(x^2-y^2)^9$$2 One parametrization of \mathscr{C} is$$ \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} ...

2

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 ...

2

Your solution is correct. And as you pointed out it implicitly assumes that $x,y$ are real. This is because your definition of the limit already did, when you used "$0 < \sqrt{x^2+y^2} < δ$". If you want a more general definition of limit that works for complex $x,y$, then it goes as follows: $\def\cc{\mathbb{C}}$ $f \colon \cc^2 \to \cc$ is ...

2

The proof is fine as-is. To answer your latter question of how to find $\delta(\varepsilon)$, you typically work backwards. Force the difference between the function and the limit to be less than $\varepsilon$, and then use algebraic manipulations (or casework) to get the difference to "look like" the distance function. In your problem, we must force ...

2

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 ...

2

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.

1

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 ...

1

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.

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 $$\int y\frac{dx}{dt}dt$$

1

The expression $\int_0^\infty \frac{\sin x}{x} dx$ is an abuse of notation, the precise meaning of this expression should be: $$\lim_{t \to \infty} \int_0^t \frac{\sin x}{x} dx = \frac{\pi}{2} \tag{1}$$ since in fact $x^{-1} \sin x$ is not Lebesgue integrable over $(0, +\infty)$, because its positive and negative parts integrate to $\infty$. To show $(1)$ ...

Only top voted, non community-wiki answers of a minimum length are eligible