I gather you're supposed to compute the integral of $G(x,y,z)=z$ over a surface $S$ that is the part of the parabolic cylinder $y=z^2$ with $0\le x\le2$ and $0\le z\le\frac{\sqrt{15}}2$ .

We can parameterize $S$ by

$\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k$

with the given constraints on $x$ and $z$ . Take the normal vector to $S$ to be

$\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k$

so that the surface element is

$\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz$

Then in the integral, we have

$\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}$

5 0

3 years ago

Jenna manipulated the equation 4x + 7=10 by adding -7 to both sides.Which of the follolwing properties justifies this manipulati

elena-14-01-66 [18.8K]

The addition property of equality justifies this. You could also say that she simply subtracted 7, and then it would be the subtraction property of equality. This is the case because if you add or subtract (or multiply or divide) the same number on both sides of an equation, the equation will still be the same.

Hope it helps!

5 0

2 years ago

A store owner purchases thirty 75-watt light bulbs and 40 fluorescent light bulbs for a total cost of $95. A second purchase, at

N76 [4]

Answer: 75-watt are 30$ fluorescent are 30$ in the second and 65$ in the first

Step-by-step explanation: i calculated thirty 75-watt and the fluorescent and got these kinda hard to explain hope i helped:)

7 0

3 years ago

0.44 in scientific notation

babunello [35]

44x10^-2
or in other words
44 times 10 to the -2nd power

3 0

3 years ago