Consider the closed region

bounded simultaneously by the paraboloid and plane, jointly denoted

. By the divergence theorem,

And since we have

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have




Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by

, we have

Parameterize

by


which would give a unit normal vector of

. However, the divergence theorem requires that the closed surface

be oriented with outward-pointing normal vectors, which means we should instead use

.
Now,



So, the flux over the paraboloid alone is
The answer would be 32.064
Answer:
x = 2
Step-by-step explanation:
The product of distances from the intersection of secants to the near and far intersections with the circle are the same. For a tangent, the near and far points of intersection with the circle are the same. This relation tells us ...
(2√3)(2√3) = x(x +4)
12 = x² +4x
16 = x² +4x +4 . . . . . add the square of half the x-coefficient to complete the square
4² = (x +2)² . . . . . . . . write as squares
4 = x +2 . . . . . . . . . . positive square root
2 = x . . . . . . . . . . . . . subtract 2
_____
<em>Alternate solution</em>
If you believe x to be an integer, you can look for factors of 12 that differ by 4.
12 = 1×12 = 2×6 = 3×4
The factors 2 and 6 differ by 4, so x=2 and x+4=6.
4000 (4x10^3) times 400 (4x10^2) is equal to 16x10^5 ( 1,600,000)
A² + b² = c².
Make a triangle with the points on a grid.
One leg is 7 units and the other is 1 unit.
7² is 49 and 1² is 1.
49 + 1 = 50
√50 ≈ 7.071.
Hope this helps!