Calculate the surface integral ∫ ∫ ∑ Z ^ 2 DS, where ∑ is the part of cylinder x ^ 2 + y ^ 2 = 4 between 0 ≤ Z ≤ 6

Calculate the surface integral ∫ ∫ ∑ Z ^ 2 DS, where ∑ is the part of cylinder x ^ 2 + y ^ 2 = 4 between 0 ≤ Z ≤ 6

We will consider the YZ surface, which will be considered as the YZ surface \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\tick (1 + Y & # 178

It is known that the focus of the parabola is on the positive half axis of the x-axis and passes through the point m (3,6). The standard equation of the parabola is obtained

Focus f (P / 2,0), focal distance P > 0. It is a parabola with right opening
The standard equation of parabola is y ^ 2 = 2px,
Through the point m (3,6), 6 ^ 2 = 2 * 3 * P, P = 6
So y ^ 2 = 12x is the result

The focus F of the parabola is on the x-axis, and the point a (m, - 3) is in the standard equation of the parabola
And | AF | = 5

Let the parabolic standard equation be y & # 178; = 4PX
Then the focus coordinate is (P, 0)
From the meaning of the title, 4pm = (- 3) &;
(m-p)²+(-3-0)²=25
The solution of these two equations is very simple
p1=1/2 m1=9/2
p2=-9/2 m2=-1/2
p3=9/2 m3=1/2
p4=-1/2 m4=-9/2
So the standard equation of parabola is
y²=6x A(9/2, -3) m=9/2
y²=-18x A(-1/2, -3) m=-1/2
y²=18x A(1/2, -3) m=1/2
y²=-6x A(-9/2,-3) m=-9/2

Parabolic standard equation with distance from focus to collimator of 6

The outer side equals six X