# Tag Info

11

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty ... 8 Use polar coordinates, We know that$$r^2 = x^2 + y^2$$So our double integral becomes$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$Now solve. EDIT I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral. 7 Yes, using polar coordinates the boundaries are:$$0 \leq r \leq R \\ 0 \leq \theta \leq 2 \pi$$Since D(R) is the disk of radius R with center at (0,0): D(R)=\{(x,y): x^2+y^2 \leq R^2\} So we have the following:$$x = r\cos \theta, y = r \sin\theta \\ \renewcommand{\intd}{\,\mathrm{d}} \intd x \intd y = r \intd r \intd \theta ...

6

For $\varepsilon>0$ denote $$X_{\varepsilon} = \Big \{ (x,y) ~|~ \exists \eta >0, \forall (h,k) \in [-\eta,\eta]^2,~ |f(x+h,y+k)-f(x,y)|< \varepsilon \Big\}.$$ Then $f$ is continuous on $\bigcap_{\varepsilon \in \mathbb{Q}_{>0}} X_{\varepsilon}$. Lemma: For any $\varepsilon>0$, and open $O$ : $X_{\varepsilon} \cap O$ has non empty interior. ...

5

If you use polar coordinates you will get the following $$\int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta$$

5

\begin{align} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy &=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy\\ &=\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}\underbrace{\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}dx\right]}_{constant \ w.r.t.y}dy\\ ...

5

The formula of the are of the surface given as a graph of the function $z=f(x,y)$ over the region $(x,y) \in D$ is $$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA$$ In this case $D$ is the disk of radius $1$ with center at $(0,0)$: $D=\{(x,y): x^2+y^2 \leq 1\}$ $$z=f(x,y)=xy$$ $$f_x=y, f_y=x$$ So, we have the following: $$A(S)=\iint_D ... 5 Aside from showing that two limits differ as you do for y=x and y=x^2-x, you can also do this in one fell swoop: Let y=x^3-x, then you get$$\lim_{(x,y)\to (0,0)}{x^2\over x^3}$$which clearly diverges to infinity. 3 By definition we have$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=0$$and the same for \frac{\partial f}{\partial y}(0,0). 3 Hint: \cos^2 \theta \cos^2 \phi +\cos^2 \theta \sin^2 \phi = \cos^2 \theta (\cos^2 \phi + \sin^2 \phi) . 3$$\int_0^{\infty} \! \!\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \! \int_0^{2\pi} \! \frac{e^{-r^2(cos^2\theta cos^2\phi +cos^2\theta sin^2\phi + sin^2\color{red}\phi) }}{\sqrt{r^2(cos^2\theta cos^2\phi +cos^2\theta sin^2\phi + sin^2\color{red}\phi)}}r^2cos\theta \, \mathrm{d}\phi\mathrm{d}\theta\mathrm{d}r$$The red bits should be \theta and$$\forall ...

3

Notice that if you apply the cofactor expansion along the first column we have a lot of zeroes so we only need to worry about $3$ multiplied by the determinant of a smaller matrix. Notice that there is another candidate column to which to apply the cofactor expansion in the smaller matrix (the column with the most zeroes). Continue this process. Notice that ...

3

This notation is just shorthand for $dx\,dy\,dz$. It's not the greatest notation in the world as far as I'm concerned but it's certainly not the worst.

3

Yes, you are using the chain rule. In cylindrical coordinates $r=r(x,y)=\sqrt{x^2+y^2}$. So, as $r$ only has a dependance on $x$ and $y$: $$\frac{\partial \phi}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial \phi}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial \phi}{\partial y}$$

3

It depends on whether you are computing a line integral or a double integral over a region. When computing a line integral, you parameterize the path with one variable. You are integrating w.r.t. that variable. However, if you are doing a double (area) integral, then you parameterize the region in two variables, and you integrate w.r.t. these variables. ...

3

Equivalently, you may go from the following way: $$A: x=0,~ 0\le y\le 2\longrightarrow\int_A=0$$ $$B: y=2,~0\le x\le4\longrightarrow \int_B=\int_0^4(2+e^{\sqrt x})dx$$ $$C:=0\le x\le4,~y=\frac{x}2\longrightarrow \int_C=\int_0^4\left(\frac{x}2+e^{\sqrt x}\right)dx+\frac{1}2\int_0^4(xe^{x^2/4})dx$$

3

As Pringoooals suggested, use Green's Theorem. The key is to integrate $x$ first, then $y$. $\displaystyle\int_C(y+e^\sqrt{x}) dx + (xe^{y^2}) dy = \iint_{\Omega}\left[\dfrac{\partial}{\partial x}(xe^{y^2}) - \dfrac{\partial}{\partial x}(y+e^{\sqrt{x}})\right]\,dx\,dy = \iint_{\Omega}(e^{y^2}-1)\,dx\,dy = \int_{0}^{2}\int_{0}^{2y}(e^{y^2}-1)\,dx\,dy = ... 3 You have a defined region to integrate about, which is$D(R)=\{(x,y) \in \mathbb{R}^{2}| x^{2}+y^{2} \leq R^{2}\}$as the disc with radius$R$. At first, you should choose an adequate parametrization for making the problem easier. Here you might use the transformation$(r, \theta) \mapsto (x, y)=(r \cos \theta, r \sin \theta)$with$r \in [0, R]$,$\theta ...

3

Hint: Let $x = r\cos \theta$ and $y = r\sin \theta$, then $$I = \int_{0}^{2\pi} \int_{0}^R r e^{-r^2} \,\mathrm{d}r \,\mathrm{d}\theta$$

2

No. Any time you define $g(a)=\text{last time something happens}$, continuity is going to be lost the moment something stops happening. Consider the quadratic polynomial $f(x) = (x-1)^2+c$. When $c\ge 0$, it is always nonnegative. When $c<0$, it changes sign as $x$ grows, until it settles at being nonnegative at $x=1+\sqrt{c}$. So, the smallest $x$ ...

2

The results about optimization of $f(x,y)$ rest on the multivariate Taylor formula: $$f(x,y) = f(a,b)+(\nabla f)(a,b) \cdot \langle x-a,y-b \rangle+ Q(a,b)(x-a,y-b) + \cdots$$ Here the quadratic term $Q(a,b)$ has the explicit form: $$Q(a,b)(h,k) = \tfrac{1}{2}\bigl(f_{xx}(a,b)h^2+2f_{xy}hk+ f_{yy}k^2\bigr)$$ where you can think $h=x-a$ and $k=y-b$. A ...

2

$\Sigma$ is the graph of the function $z = xe^y$. That is $(x,y,z) = (x,y, \varphi(x,y)) = G(x,y)$. So $$\int_{\Sigma}\mathbf{f}\cdot d\sigma = \int_{x}\int_{y}f(G(x,y))\cdot \left(\frac{\partial G}{\partial x}\times \frac{\partial G}{\partial y}\right)dx dy$$ What should $\varphi$ be? Once you have determined that and the bounds ...

2

In answer to my own question, the equality can be shown as follows. First, we realize that $\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left (\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}\right )}{\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}} dr_1dr_2d\theta\\=\frac{1}{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left ... 2 Hint for A: Area of the region$R$could be: $$\text{Area of} ~~R=\int_RdA=\int_{z=0}^{2}\int_{\theta=0}^{2\pi}\int_{r=0}^{3}rdrd\theta dz=18\pi$$ Now try to solve $$\int_R T(x,y,z)dA=\int_{z=0}^{2}\int_{\theta=0}^{2\pi}\int_{r=0}^{3}\frac{z}{1+r^2}rdrd\theta$$ 2 You don't really need to setup any parametrization first. You should first simplify the integral and then try either$x$or$y$as your parametrization unless there are other obvious choices.$\displaystyle\;\int_C y dx$is just negative of the area of$R$and hence$= -\frac12 (2 \times 4) = -4$.$e^{\sqrt{x}} dx$is a total differential, so its integral ... 2 The general gaussian integration formula is $$\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty} e^{-\frac12 x^T Ax}dx_1\ldots dx_n=\frac{(2\pi)^{\frac{n}{2}}}{\sqrt{\operatorname{det}A}}.$$ In your case,$n=2$and$A=\left(\begin{array}{cc} 1 & -1/2 \\ -1/2 & 1\end{array}\right)$. 2 The paraboloid and cylinder intersect at$x=9$, so the height of the paraboloid is$h=9$then find the surface area by integrating: $$A=\int\int \sqrt{1+(2y)^2+(2z)^2}$$ which you need to convert to polar coordinates... 2 The hard way is to compute$A(R) = \displaystyle\iint\limits_{R}\,dx\,dy = \int_{0}^{1}\int_{0}^{x}\,dy\,dx$. The easy way is to note that$R$is a right triangle with base length$1$and height$1$. 2 Your answer is correct, but not in the form they expected. It looks like you have$33\sqrt{1/11}, but \begin{align*}&=3*11*\sqrt{1/11} \\ &= 3 \sqrt{11^2/11} \\ &= 3 \sqrt{11} \end{align*} 2 The tangent vectors to the surface are $$\mathbf{t}_u=\left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right)\\ \mathbf{t}_v=\left(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}\right)$$ The normal to the surface is\mathbf{n}=\mathbf{t}_u\times\mathbf{t}_v\$. The ...

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