3

We can do it by hand. We have $f(0,0) = 0$, so we mush show on some lines through $(0,0)$, $f$ is positive in both directions, and on other lines it is negative in both directions. For small $t$ we have $$ f(t,0) = 2t^2-t^4 \sim 2t^2 > 0 $$ on both sides of $(0,0)$, and for small $t$ we have $$ f(t,t) = -2t^4 < 0 $$ on both sides of $(0,0)$.


3

The integral is equal to $$ \int_S 1 = \text{Surface area}=4\pi $$ And to continue your approach you can easily see that $f(r(u,v)) =1$ thus the integral becomes $$ \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} |\cos(v)| dudv = 4\pi \times \int_{0}^{\pi/2} \cos(v) dv = 4\pi $$


2

The Divergence Theorem is absolutely the right approach, as all the hypotheses of the theorem hold. You should always prefer to do a volume integral rather than a surface integral. In this case, you might also recognize that \begin{multline*} \iiint_K x\,dV = \bar x\, \text{volume}(K), \quad \iiint_K y\,dV = \bar y\, \text{volume}(K), \quad\text{and} \\ \...


2

Your region is $R: (x-1)^2+(y+1)^2+(z-2)^2 \leq 4$. If you are applying divergence theorem, you need to evaluate $\displaystyle \iiint_R 2(x+y+z) \ dV$ Just to simplify your working, consider the below substitution - $u = x-1, v = y+1, w = z-2$ and region $R$ transforms to $E: u^2+v^2+w^2 \leq 4$ which is a sphere of radius $2$ centered at the origin. ...


1

Ok, lets look at the general definition of a differential $df$ on normed spaces, for $f:\mathbb{S}\to\mathbb{P}$ $$f(x+h)=f(x)+df(x,h)+o(||h||_S)$$ where the differential $df(x,h)$ is linear and continuous in $h$. Lets look at $\mathbb{S}=\mathbb{R}^m,\mathbb{P}=\mathbb{R}^n$. Then any linear operator $\mathbb{R}^m\to\mathbb{R}^n$ can be represented by some ...


1

After some time, I realized that I made a mistake in my initial attempt. With more care, it looks like the Gaussian solution satisfies equation (1), but I am unsure if this is correct. So suppose that the substitution of $\tilde{t} = t^a$ is correct and equation (1) of the question can be reduced to $$\frac{\partial{f}}{\partial{\tilde{t}}}=\frac{\partial^2{...


1

As per Stokes' theorem, $\displaystyle \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \int_C \vec {F} \cdot dr \ $, where $C$ is the boundary curve of the surface $S$. $S$ is $x^2+y^2+z^2 = 4, z \geq 0$. Its boundary curve is circle $x^2+y^2 = 4$ at $z = 0$. Parametrize the boundary curve in polar coordinates as $r(t) = (2 \cos t, 2 \sin t, 0), 0 \...


1

Morse theory says that every Morse function $f$ (all critical points were non-degenerate and distinct critical points take distinct critical values.) $\#\min+\#\max-\#\mathrm{saddle}=\chi(M)$. So, in the case of torus (that its Euler char is $0$), the functions must have $\#\mathrm{saddle}-\#\max=2$. So if it has no max point, then the number of saddle ...


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