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4

Your definitions of $r$ at the beginning are getting you into trouble. Bad things happen at $\sin \theta = 0$ or $\cos \theta = 0$. Try $r = \sqrt{x^2+y^2}$. Then $$\frac{\partial{r}}{\partial{x}} = \frac{x}{r}; \frac{\partial{r}}{\partial{y}} = \frac{y}{r}.$$ Then, using Fantini's $\theta = \tan ^{-1}(\frac{y}{x})$, we get $$\frac{\partial ... 4 If all the roots are real, then it is proven. Otherwise, there is at least one non-real root z. But then \bar{z}, the conjugate, is also a root (can you show this?). Thus, there are an even number of non-real roots. Since there are three roots in \mathbb{C}, it follows that at least one must be real. 4 Indeed, cause of the symmetric, the final triple integrals are as:$$V=8\left(\int_{\theta=0}^{\pi/4}\int_{r=0}^{\sec(\theta)}\int_{z=0}^{\sqrt{1-r^2\sin^2(\theta)}}+\int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{\csc(\theta)}\int_{z=0}^{\sqrt{1-r^2\cos^2(\theta)}}\right)$$But I don't think this way of using the Cylindrical Coordinates make the integrals above ... 3 The equation of the line can be written as:$$\frac{x-1}{-2}=\frac{y+1}{4}=\frac{z-3}{-2}=t$$which tells us that one possible point on the desired plan would be$$A:~(-1,+1,-3)$$and also the leading vector of the line is:$$\vec{w}:=(-2,4,-2)$$On the other hand, the normal vector of the given plan is \vec{u}:= (2,2,0). Now find the vector$$\vec{n}:= ...

3

And an even simpler method for finding this integral is replacing $x$ with $u^2$ , $y$ with $v^2$ and $z$ with $w^2$. Instead of this : $$\int_{0}^1\int_{0}^{\left(1-\sqrt x\right)^2}\int_{0}^{\left(1-\sqrt x -\sqrt y\right)^2}\,dz\,dy\,dx$$ $dx = 2u\,du\\dy=2v\,dv\\dz=2z\,dz$ you will get this : ...

3

All odd polynomials have at least one real root because of the intermediate value theorem. To prove this just plug in a very large positive number and a very large negative number for $x$ (e.g. $10^{23}$ and $-10^{23})$ and note that corresponding $y$ values will have opposite signs. Then the IVT tells you that there is at least one value of $x$ between the ...

3

Sounds like Chain rule. First, notice that $$2^{\sqrt{t}} = \left(e^{\log 2}\right)^{\sqrt{t}} = e^{\log 2\sqrt{t}}$$ Then $$\frac{dy}{dt} = \frac{d}{dt}\left(e^{\log 2\sqrt{t}}\right) = \left(e^{\log 2\sqrt{t}}\right)\frac{d}{dt}\left(\log 2\sqrt{t}\right) = 2^\sqrt{t}\frac{\log 2}{2\sqrt{t}}$$ Where $\log$ is the natural logarithm.

3

Your formulas establish that $g$ is $C^1$ away from the origin, but establish nothing at the origin. (Your continuity claim for $x/\sqrt{x^2+y^2}$ certainly fails at the origin.) Check the definition and ask if there is a linear map $Dg(0)\colon \Bbb R^2\to \Bbb R$ with the appropriate properties. EDIT: To wit, we need $$\lim_{h,k\to 0} ... 2 You want to convey information in higher dimensions on a 2D screen. What is generally used for 3D, is a projection of a two dimensional surface, where z is a function of x,y . For 4D color is generally added, so that in addition to height, a color is used to represent the value in the fourth dimension. Other methods used include drawing multiple ... 2 Noticing antisymmetry about the line y=-x is the "best" solution, but here is another approach if you know Stoke's theorem:$$\iint_D 2xe^{x^2+y^2} -2ye^{x^2+y^2} dydx = \iint_D d(e^{x^2+y^2}dx +e^{x^2+y^2}dy) =\int_{bD}e^{x^2+y^2}dx +e^{x^2+y^2}dy$$But x^2+y^2 = 4 on bD= circle of radius 4, so$$=\int_{bD}e^4dx+e^4dy$$Now compute ... 2 To verify a vector field is conservative or not, use:$$\nabla \times F = 0$$or say$$\begin{vmatrix} \frac{\partial}{\partial x}& \frac{\partial}{\partial y} \\ M& N \\\end{vmatrix} = 0$$In this case, after my calculation, it is indeed conservative. 2 Well, by a straightforward computation,$$ \alpha \wedge d\alpha = (\omega - f^{-1}dx^{m+1}) \wedge d (\omega - f^{-1}dx^{m+1})\\ = (\omega - f^{-1}dx^{m+1}) \wedge (d\omega + f^{-2} df \wedge dx^{m+1})\\ = \omega \wedge d \omega + \omega \wedge f^{-2}df \wedge dx^{m+1} - f^{-1}dx^{m+1} \wedge d\omega\\ = \omega \wedge d\omega - f^{-2} (df \wedge \omega ...

2

You are not integrating the region $R$. The idea of the exercise is to deform $R$ into a square with sides parallel to the axis in the plane by a counterclockwise rotation of an angle equal to $\frac{\pi}{4}$. The required transformation is then $$(x,y)\mapsto (u,v)=\frac{\sqrt{2}}{2}(x-y,x+y).$$ To finish the proof you need the formula for the change of ...

2

Look at this plot in $1/8$ of all space, $x,y,z\ge0$: We have to find two volumes in this part of $\mathbb R^3$. Once, you get the whole volume of the following volume inside the sphere, you can multiply it by $8$ and then subtract it from $4/3\pi(2a)^3$. But about this below shape: $$V_1: \phi|_0^{\pi/6}, ~~\theta|_0^{\pi/2},~~\rho|_0^{2a}\\ V_2: ... 2 Your surface is defined by a polynomial equation f(x,y,z) = 0. You could interpret this situation geometrically in the following way: f:R^3\to R gives a color to every point of the 3D space (imagine you have a color palette indexed by real numbers). Assume white is indexed at 0, then your surface is the locus of points with white color. The ... 2 You really do need to know the definition in order to figure this problem out. That is, the limit definition:$$ \dfrac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \dfrac{f(h,0) - f(0,0)}{h} = \lim_{h\to 0} \dfrac{\dfrac{h \cdot 0}{h^2+0} - 0}{h} = \lim_{h\to 0}\dfrac{0}{h} = 0 $$and so the partial derivative with respect to x is zero. Similarly,$$ ...

2

It seems that the answer is $|B_r| = LV_r$, where $L$ is the length of the curve $\gamma$ and $V_r$ is the volume of the ball with radius $r$ in $\mathbb R^{n-1}$. This is a bit odd to me because $B_r$ is not isometric to the cylinder, so before the calculation I thought I would get something related to the curvature of $\gamma$. The point is to find a ...

1

No, $a_i$ is a scalar valued function that takes a vector among its arguments. Just because the input is a vector, the output need not be. But the collection of all $a_i$, $(a_1,\dots,a_n)$, can be considered as a vector-valued function, and often it is convenient to do so. Then the PDE takes the form $$-\operatorname{div} A(x,u(x),\nabla u(x)) + ... 1$$\nabla[\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x})]$$is not well defined as \mathbf{f}:\mathbb{R}^m\rightarrow\mathbb{R}^n, \mathbf{g}:\mathbb{R}^p\rightarrow\mathbb{R}^m and x\in \mathbb{R}^p (in other words, \mathbf{f}(\mathbf{x}) is not well defined if p\neq m). On the other hand, ... 1 Summarizing the extend discussion (sorry!) in the above comments, You cannot apply Stoke's theorem repeatedly to reduce the dimension of the region of integration over and over, because the boundary of the boundary of set is empty. For example, the boundary of the solid ball is the sphere, but the sphere has no boundary, so we have to stop there. ... 1 \delta(x) is not really a function in classical sense. For the purpose of deriving an expression without involving the concept of distribution, we will treat it as some sort of derivative of a step function. Assume all k_i \ne 0, let$$\lambda_i = |k_i|,\quad y_i = \begin{cases}x_i,& k_i > 0\\1-x_i,& k_i < 0\end{cases}, \quad K = ...

1

OK, so I think i worked it out now. If we use the Physics/ Einstein notation we can write: $$\det(\nabla u)= \varepsilon_{i_1,\cdots,i_n}\partial_{i_1} u^1\cdots\partial_{i_n} u^n$$ where $\varepsilon_{i_m,\cdots,i_m}=0$ and $\varepsilon$ is anti symmetric considered as an n-linear form. Writing this equation more explicitly in the first summed component we ...

1

This is not quite right. The region of integration there is not curved but straight like this: Instead, maybe to Also please look at this question here: Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer).

1

Note that $\cos(\phi-\pi)=-\cos(\phi), \cos(\phi - \pi/2)=\sin(\phi)$ so $A_3=A_c+D\theta \cos(\phi)$ which gives $A_1+A_3=2A_c=A_2+A_4$ If these are not satisfied your equations are contradictory. You can then write $$A_3-A_1=2D\theta \cos(\phi)\\A_4-A_2=2D\theta \sin(\phi)\\ \phi=\arctan\left(\frac {A_4-A_2}{A_3-A_1}\right)\\ \theta=\frac ... 1 Basically, for any function f(x, y, z) and vector field \mathbf V(x, y, z), we have \nabla \times (f \mathbf V) = \nabla f \times \mathbf V + f \nabla \times \mathbf V; \tag{1} in the present case, taking \mathbf V = \vec r, it is easy to see by direct calculation that \nabla \times \mathbf V = \nabla \times \vec r = 0, \tag{2} and since ... 1 OK, here's the structure of the calculation Step 1: find a linear map which maps the tetrahedron to something easy to parametrize. Clearly a linear map can only take 0 to 0, so we can only play with the three other vectors. We use the linear map defined by (1,2,3)\mapsto(1,0,0), (0,1,2)\mapsto(0,1,0) and (-1,1,1)\mapsto(0,0,1). Since these three ... 1 Proof by contradiction Assume ||\nabla f||\geq 2 for each point in D and define a vector field v on D such that$$\frac{dx}{dt}=\nabla f$$Consider the solution curve \gamma(t) passing through the origin at time 0, where t is natural parameter (||\frac{d\gamma}{dt}||=1. Since ||\nabla f||>0 for each point in D, the curve cannot ... 1 1) Evaluation of I_{\max}(k,n) It is clear that \max x = x so I_{\max}(1,n)=\int\limits_0^1 x_1^n\,dx_1=\frac{1}{n+1}. Also it is obvious that \max\limits_{1\le i\le k}x_i=\max\left(x_k,\max\limits_{1\le i\le k-1}x_i\right). Then ... 1 I know this answer is incredibly delayed; but just to sumarise the last post. If I gave you the function$$ f(x,y) = sin(x)+3y^2$$And asked you for the partial derivative with respect to x, you should write:$$ \frac{\partial f(x,y)}{\partial x} = cos(x)+0 Since y is effectively a constant with respect to x. In other words; substituting a value for ...

1

Having a non-zero differential is not sufficient to get what you want. If you assume that the differential of $f$ at $0$ has rank $2$, then the local inversion theorem says that the map $f$ is a local diffeomorphism in a neighborhhod of $0$. In particular, there exists a neighborhood $U$ of $(0,0)$ such that if $(x,y) \in U$ with $f(x,y) = f(0,0)$ then ...

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