Datrys x^2+y^2+z^2=8 | Microsoft Math Solver

Datrys ar gyfer x (complex solution)

x=-\sqrt{8-z^{2}-y^{2}}

x=\sqrt{8-z^{2}-y^{2}}

Datrys ar gyfer y (complex solution)

y=-\sqrt{8-z^{2}-x^{2}}

y=\sqrt{8-z^{2}-x^{2}}

Datrys ar gyfer x

x=\sqrt{8-z^{2}-y^{2}}

x=-\sqrt{8-z^{2}-y^{2}}\text{, }|y|\leq \sqrt{8-z^{2}}\text{ and }|z|\leq 2\sqrt{2}

Datrys ar gyfer y

y=\sqrt{8-z^{2}-x^{2}}

y=-\sqrt{8-z^{2}-x^{2}}\text{, }|x|\leq \sqrt{8-z^{2}}\text{ and }|z|\leq 2\sqrt{2}

Cwis

Algebra

x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 8

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Probability that x^2+y^2+z^2=0 mod p

https://math.stackexchange.com/questions/2164308/probability-that-x2y2z2-0-mod-p

Found it. Given odd dimension n and quadratic form f = a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2, everything in a finite field with odd number of elements q, the count \#\left(f = b\right) \; = \; q^{n-1} + q^{(n-1)/2} \; \; \chi \left( \; (-1)^{(n-1)/2} \; b a_1 a_2 \ldots a_n\right). ...

Find the volume of the region enclosed by x^2+y^2+z^2=2 and x^2+y^2=z.

https://math.stackexchange.com/q/1806866

I can't understand why this solid is getting chopped up into two pieces. We know the \theta goes all the way around, from 0 to 2\pi, then the boundary conditions can be solved as z=r^2, r^2+z^2=r^2+r^4=2 ...

(x+2y+z)\cdot \left( \frac{x}{y} +\frac{2y}{z}+\frac{z}{x}\right) > 12 for x^2+y^2+z^2=3

https://math.stackexchange.com/q/1805032

Let x=u^2,\quad y=v^2,\quad z=w^2,\qquad(1) then the inequality becomes (u^2+2v^2+w^2)\left(\dfrac{u^2}{v^2}+\dfrac{2v^2}{w^2}+\dfrac{w^2}{u^2}\right) > 12 for u^4+v^4+w^4=3. Using ...

Volume of the solid bounded by the sphere x^2 + y^2 + z^2 = 9 and paraboloid 8z = x^2 + y^2

https://math.stackexchange.com/q/1050273

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For which a>0 does the equation x^2+y^2+z^2=a have a solution in \mathbb{Q}_2?

https://math.stackexchange.com/questions/1126803/for-which-a0-does-the-equation-x2y2z2-a-have-a-solution-in-mathbbq

With ordinary integers: Theorem of Gauss that x^2 + y^2 + z^2 integrally represents all positive integers except those 4^k (8 n+7). Meanwhile, if x^2 + y^2 + z^2 \equiv 0 \pmod 4, then x,y,z ...

Area of smaller part of surface

https://math.stackexchange.com/q/1896215

Something is your with the integrand. y_x=-\frac{x}{\sqrt{8-r^{2}}}, \quad y_z=-\frac{z}{\sqrt{8-r^{2}}}\Rightarrow y_{x}^{2}+y_{z}^{2}+1=\frac{r^2}{8-r^{2}}+1=\frac{8}{8-r^{2}}. Hence S=\int\int\sqrt{y_{x}^{2}+y_{z}^{2}+1}=\int_{0}^{2\pi}\int_{0}^{2}r\sqrt{\frac{8}{8-r^{2}}} dr d\theta\\ =-\pi\sqrt{8}\int_{0}^{2}\frac{d(8-r^2)}{\sqrt{8-r^{2}}} =-\pi\sqrt{8}\left[2\sqrt{8-r^{2}}\right]_{0}^{2} =8\pi(2-\sqrt{2}). ...

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