Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

7 0

3 years ago

Please help!! This is so confusing! 100 points!!

masya89 [10]

Answer:

The value of the expression given is:

Step-by-step explanation:

First, you must divide the expression in three, and to the final, you can multiply it:

[(3^8)*(2^(-5))*(9^0)]^(-2)
[(2^(-2))/(3^3)]^4
3^28

Now, we can solve each part one by one:

First part.

3^8 = 6561
2^(-5) = 0.03125
9^0 = 1 (Whatever number elevated to 0, its value is 1)
(6561 * 0,03125 * 1) = 205.03125

And we elevate this to -2:

205.03125^-2 = 2.378810688*10^(-5) or 0.00002378810688

Second part.

2^(-2) = 0.25
3^3 = 27
0.25 / 27 = 9.259259259 * 10^(-3) or 0.00925925925925

And we elevate this to 4:

0.00925925925925^4 = 7.350298528 * 10^(-9) or 0.000000007350298528

Third Part.

3^28 = 2.287679245 * 10^13 or 22876792450000

At last, we multiply all the results obtained:

0.00002378810688 * 0.000000007350298528 * 22876792450000 = 3.999999999999999999 approximately 4

We approximate the value because the difference to 4 is minimal, which could be obtained if we use all the decimals in each result.

3 0

3 years ago

Read 2 more answers

Solomon graphs a point on the coordinate plane. Which point did Solomon graph?

Mkey [24]

Answer:

Try your best! I know you can do it!

5 0

3 years ago