Find the volume of solid surrounded by z = x ^ 2 + y ^ 2 and z = 2-radical (x ^ 2 + y ^ 2)

Find the volume of solid surrounded by z = x ^ 2 + y ^ 2 and z = 2-radical (x ^ 2 + y ^ 2)

x² + y² = zz = 2 - √(x² + y²) ==> √(x² + y²) = 2 - z ==> x² + y² = (2 - z)² = 4 - 4z + z²z = 4 - 4z + z² ==> z² - 5z + 4 = 0 ==> (z - 4)(z...

The first problem is to find the volume of curved surface z = x & # 178; + 2Y & # 178; and z = 6-2x & # 178; - Y & # 178

From z = x & # 178; + 2Y & # 178; and z = 6-2x & # 178; - Y & # 178;, we can get that the intersection line of the two is X & # 178; + Y & # 178; = 2
V=∫∫[(x²+2y²)-(6-2x²-y²)]ds
The integral domain is d = {(x, y) | X & # 178; + Y & # 178; = 2}

Please refer to a higher integral problem, ∫∫ XYD σ, D: 0 ≤ y ≤ 1, &# 189; Y & # 178; ≤ x ≤ √ 3-y & # 178;, is the root sign 3-y & # 178;

∫∫xydσ=∫(0,1)ydy∫（½y²,√3-y²）xdx=∫（0,1）y*1/2x²|(½y²,√3-y²)dy=1/2∫（0,1）（3y-y³-y^5/4）dy=1/2*[3y²/2-y^4/4-y^6/24]|(0,1)=1/2*(3/2-1/4-1/24)=29/...

Find the area of the square of 4-x-y contained in the square of cylinder x + the square of y = 2x under the root sign of Z = upper hemisphere

The center of the projection plane of the upper hemisphere z = √ [4-x ^ 2-y ^ 2] is (0,0), and the radius is 2
Cylinder x ^ 2 + y ^ 2 = 2x, that is, (x-1) ^ 2 + y ^ 2 = 1, the center of the cross section is (1,0), and the radius is 1
Surface equation z = √ [4-x ^ 2-y ^ 2]
Surface area A1 = ∫∫ Da = ∫∫ √ (1 + P ^ 2 + Q ^ 2) DXDY
Zone D1: 0 ≤ x ≤ 2,0 ≤ y ≤ 1
Then the area is 2 times of A1, that is, a = 2A1 = 2 ∫ ∫ √ (1 + P ^ 2 + Q ^ 2) DXDY
P=∂z/∂x=-x/√[4-x^2-y^2],Q=∂z/∂y=-y/√[4-x^2-y^2]
1+P^2+Q^2=1+(x^2+y^2)/(4-x^2-y^2)=4/(4-x^2-y^2)
dA=√(1+P^2+Q^2)dxdy=2/√[4-x^2-y^2]*dxdy
A=2∫∫dA=∫∫2/√[4-x^2-y^2]*dxdy (0≤x≤2,0≤y≤1)
=2 ∫ ∫ 2 / √ [4-R ^ 2] * rdrd θ (polar coordinate: 0 ≤ R ≤ 2cos θ, 0 ≤ θ ≤ π / 2)
=2∫dθ∫2/√[4-r^2]*rdr
=2∫4(1-sinθ)dθ
=8(θ+cosθ)
=8(π/2-1)
=4(π-2)

If the intercept of the straight line L on the two coordinate axes is equal and tangent to the circle C: X & sup2; + (Y-2) & sup2; = 1, then the equation of L is

Circle: X & sup2; + (Y-2) & sup2; = 2. Center of circle (0,2), radius r = √ 2. (1) when the intercept is not 0, let the tangent equation be (x / a) + (Y / b) = 1. (|a | = |b | ≠ 0). It is easy to know that the distance from the tangent to the center of circle (0,2) is √ 2, |2a-ab | / √ (A & sup2; + B & sup2;) = √ 2. = = = = > |2a-ab | = 2 | a |. = = = > |b-2 | = 2

Through a (1,3), we can get the following results: 1. The equation of line with equal intercept of line L on two coordinate axes; 2. The relation of circle x2-6x + Y2 + 2Y = 0 to symmetrical circle of line OA
Given that the line L passes through a (1,3), we can find: 1. The equation of line L with equal intercept on two coordinate axes; 2. The equation of circle x2-6x + Y2 + 2Y = 0 with respect to the symmetrical circle of line OA

If the line L is y = x + B, and the point a (1,3), we can get b = 4, that is y = - x + 4. If we cross the origin, we can know that y = 3x2, we can know that the coordinates of the center of the circle B are (3, - 1), and the center of the circle is symmetric to the line OA, and the point C is (x, y). Then there is a point ((3 + x) / 2, (Y-1) / 2) on the line OA, y = 3x, that is, (Y-1) / 2 = 3 (3 + x)

Through point P (2,0), make the tangent of circle C: x ^ 2 + y ^ 2-6x-4y + 12 = 0, and find the tangent equation

Wrong plan!
Given the equation x & # 178; + Y & # 178; - 6x - 4Y + 12 = 0
It is reduced to: (x - 3) &# 178; + (Y - 2) &# 178; = 1
The center of the circle (3,2) is obtained and the radius is 1
(when k exists)
Because the tangent goes through P (2,0),
Let the slope of the tangent be K,
Then the equation of tangent is Y - 0 = K (x - 2) (point oblique)
KX - Y - 2K = 0
Using the formula of distance from point to line:
There is ∣ k * 3 - 2 - 2K ∣ / [√ (K & # 178; + 1)] = 1 (the distance from the center of the circle to the tangent is the radius)
It is reduced to ∣ K - 2 ∣ = √ (K & # 178; + 1)
The square of both sides gives K & # 178; - 4K + 4 = K & # 178; + 1
So k = 3 / 4
So the tangent equation is y = (3 / 4) (x - 2)
We get 3x - 4Y - 6 = 0
(when K does not exist)
The tangent is just perpendicular to the X axis,
And because P (2,0) is too high,
So the tangent equation is x = 2
So the tangent equation is 3x - 4Y - 6 = 0 or x = 2
Note: there must be two tangents passing through a point outside the circle, so when calculating the tangent equation, there must be two solutions
But if there is only one solution to K in the oblique formula of tangent, then another is necessary to discuss (as above)
Of course, if you draw a picture before solving the problem, it will be clearer

A point m (2,3) outside the circle x ^ 2 + y ^ 2 = 1. Make two tangents Ma and MB of the circle, and the tangents are a and B respectively. Find the equation ab of the straight line

Let m (2,3), O (0,0),
MA^2=OM^2-r^2=13-1=12
Therefore, the equation of the circle m passing through AB with m as the center is as follows:
（x-2)^2+(y-3)^2=12
Expanded x ^ 2 + y ^ 2-4x-6y + 1 = 0
The subtraction of two circles is a straight line AB:
2x+3y-1=0

Find a point m (2,3) outside the circle x ^ 2 + y ^ 2 = 1, make two tangent lines Ma and MB of the circle, and the tangent points are a and B respectively, and find the equation of straight line ab
The center of x ^ 2 + y ^ 2 = 1 is O (0,0)
Because OA is vertical to Ma and ob is vertical to MB, O, a, m and B are all round
O. The common circle of a, m and B is the circumcircle of APB
The diameter of the circle is OM
|Om | = √ (4 + 9) = √ 13 -- is this place right?

A very simple result: circle x ^ 2 + y ^ 2 = R ^ 2 outside a point n (a, b), make the circle of two tangent Na, Nb, tangent points are a, B, straight line AB equation is: ax + by = R ^ 2, for this problem: r = 1, a = 2, B = 3, can be substituted. Note the previous results below: let a (x1, Y1), B (X2, Y2) through a tangent L1: X1 * x + Y1 *

Given that the circle x ^ 2 + y ^ 2 = 1 and the circle (x-3) ^ 2 + (y-4) ^ 2 = 4, the tangent lengths Ma and MB of the two circles from point m (x.y) are equal, the trajectory equation of M is obtained

The square of the distance between M and the two centers is respectively x ^ 2 + y ^ 2 = ma ^ 2 + 1, (x-3) ^ 2 + (y-4) ^ 2 = MB ^ 2 + 4. The difference between the two equations is 3x + 4y-11 = 0