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muminat
3 years ago
7

Verify that the Divergence Theorem is true for the vector field F on the region E. F(x, y, z) = z, y, x , E is the solid ball x2

+ y2 + z2 ≤ 64 For your answer, put in the flux across the boundary of E with respect to the outward normal.
Mathematics
1 answer:
Setler [38]3 years ago
8 0

\vec F(x,y,z)=\langle z,y,x\rangle

has divergence 1, so by the divergence theorem, the flux of \vec F across the boundary of E is exactly the volume of E,

\displaystyle\iiint_E\mathrm dV=\int_0^\pi\int_0^{2\pi}\int_0^8\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

or simply \dfrac43\pi8^3=\boxed{\frac{2048\pi}3}.

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Answer:

0.4060

Step-by-step explanation:

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3 years ago
What is the simplest form of ^4 sqrt 81x^8y^5
lara31 [8.8K]

ANSWER

3 {x}^{2} y \sqrt[4]{y}

EXPLANATION

We want to simplify:

\sqrt[4]{81 {x}^{8} {y}^{5}  }

We can split the radical sign to obtain:

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{5} }

Or

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}  \times y}

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

\sqrt[4]{ {3}^{4} }  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

Recall that:

\sqrt[n]{ {a}^{m} }  =  {a}^{ \frac{m}{n} }

{3}^{4 \times  \frac{1}{4} }  \times  {x}^{8 \times  \frac{1}{4}  }  \times  {y}^{4 \times  \frac{1}{4} }\times  \sqrt[4]{y}

3 {x}^{2} y \sqrt[4]{y}

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