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Is the hyperbolic paraboloid obtained by y = ax ^ 2 rotating around the X axis? But how to deduce the hyperbolic paraboloid equation? If y = ax ^ 2 rotates around the X axis, we should get √ y ^ 2 + Z ^ 2 = ax ^ 2. Can this equation change to hyperbolic parabolic equation?
The equation of hyperbolic paraboloid, such as x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = Z, can't be a surface of revolution, because no matter how formulated, it can't be the sum of squares
Calculate the multiple integral ∫ ∫ √ X & # 178; + Y & # 178; D σ, where the integral region D is the common part of X & # 178; + y square ≥ 1 and X & # 178; + Y & # 178; ≤ 4
∫∫√x²+y²dσ
=∫∫p·pdpdθ
=∫(0,2π)dθ∫(1,2)p²dp
=2π p³/3 |(1,2)
=2π·(8-1)/3
=14π/3
The integral is calculated as follows: (1) I = ∫ (d) YD σ, and the integral domain D is bounded by the curves Y & # 178; = x and y = - x + 2
(2) The Euler integral I = ∫ e ^ (- X & # 178;) DX is obtained by using the double integral in polar coordinates, where is the upper limit and lower limit of the integral
1I = ∫ (d) YD σ = ∫ (- 2 - > 1) YDY ∫ (y ^ 2 - > 2-y) DX = - 9 / 42 ∫ (0 - > ∞) e ^ (- x ^ 2) DX = ∫ (0 - > ∞) e ^ (- y ^ 2) dy, so ∫ (0 - > ∞) e ^ (- x ^ 2) DX = √ [∫ (0 - > ∞) e ^ (- x ^ 2) DX * ∫ (0 - > ∞) e ^ (- y ^ 2) dy] = √ [∫ (0 - > ∞) ∫ (0 - > ∞) e ^ (- x ^ 2-y ^ 2)
A = {x | y = radical 1-x & # 178;, X ∈ Z}. B = {y | y = x & # 178; + 1, X ∈ a}, find a ∩ B=
It's better to have a detailed process that I can understand. I really can't understand this problem
Y = √ (1-x & # 178;) gives 1-x & # 178; ≥ 0 gives - 1 ≤ x ≤ 1 x ∈ Z, so x takes - 1,0,1
A = {- 1,0,1} B = {y | y = x & # 178; + 1, X ∈ a} so when x is - 1,0,1, y is 2,1,2
B={2,1}
A∩B={1}
If there is a point P passing through the straight line y = x and a tangent line is drawn from the circle x square + y square - 6x + 7 = 0, then the minimum value of the tangent line? (detailed explanation)
(x-3)²+y²=2
The radius is fixed
So if the tangent is the shortest, then p is the shortest to the center of the circle (3,0)
So, let's make a vertical line y = x through the center of the circle, and P is perpendicular to the foot
Y = x, the slope of the vertical is - 1
So it's x + Y-3 = 0
And y = x are (3 / 2,3 / 2)
So p (3 / 2,3 / 2)
The number of lines tangent to the circle C: x ^ 2 + (y + 5) = 9 with equal intercept on the two coordinate axes is
The number of lines tangent to the circle C: x ^ 2 + (y + 5) ^ 2 = 9 and with equal intercept on the two coordinate axes is 4
Slope of line with equal intercept = ± 1
That is to say, the included angle with X axis is 45 degrees
Circle C: the angle between the circumscribed square edge and the x-axis is 45 degrees
(or circumscribed square with an angle of 0 ° between the edge and the x-axis, rotated 45 °)
The four sides of a square are the straight lines
Tangent equation
y=x+3√2-5
y=x-3√2-5
y=-x+3√2-5
y=-x-3√2-5
2. To know that the circle is tangent to the Y axis, the center of the circle is in the straight line x-3y = 0, and the circle passes through the point a (6,1), the circular equation is solved
1. Let the straight line be x / A + Y / a = 1 and substitute into P: 1 / A + 2 / a = 1, then: a = 3, so the straight line is x + y = 32. Let the center of the circle be (3a, a), which is tangent to the Y axis, then r = | 3a | that is, the circle is (x-3a) &# 178; + (Y-A) &# 178; = (3a) &# 178; substitution point (6,1): (6-3a) &# 178; + (1-A) &# 178; = 9A & # 178; simplification: a & # 178; - 38a +
Find a linear equation which is tangent to circle (X-2) ^ 2 + (y + 2) ^ 2 = 1 and has equal intercept on X, Y axis!
There are two cases: when the distance from the origin y = KX (2, - 2) to the straight line is equal to radius = 1 (2k + 2) / root sign (k ^ 2 + 1) = 1, we can get 3K ^ 2 + 8K + 3 = 0k = (- 4 - root sign 7) / 3 or K = (- 4 + root sign 7) / 3. However, the slope of the origin must be - 1. Let the linear equation be x + y + C = 0 (2, - 2) and the absolute value of the distance from the straight line 1C = root sign 2
Find the equation of a line which is tangent to the square of circle x + the square of circle Y - 4x + 2 = 0 and has the same intercept on X and Y axes
(X-2) & sup2; + (y + 1) & sup2; = 5 Center (2, - 1), radius √ 5, if passing through the origin, is kx-y = 0, the distance from the center of the circle to the tangent is equal to the radius, so | 2K + 1 | / √ (K & sup2; + 1) = √ 5 square 4K & sup2; + 4K + 1 = 5K & sup2; + 5K & sup2; - 4K + 4 = 0k = 2, but the origin x + y + a = 0, so | 2-1 + a | / √ (1 & sup2; + 1 & sup2
A linear equation which is tangent to the square of the circle C: x plus the square of (y + 5) = 3 and has the same intercept on the x-axis and y-axis
X ^ 2 + (y + 5) ^ 2 = 3, i.e. the center O (0, - 5), radius = √ 3, the intercept of axis and the transverse intercept are opposite to each other, so let the straight line be X-Y + B = 0. Because the distance from the center O (0, - 5) to the straight line X-Y + B = 0 = √ 3, so | 5 + B | / √ 2 = √ 3, i.e. B ^ 2 + 10B + 19 = 0, B = - 5 + √ 6, and B = - 5 - √ 6
RELATED INFORMATIONS
- 1. Find the volume of solid surrounded by z = x ^ 2 + y ^ 2 and z = 2-radical (x ^ 2 + y ^ 2)
- 2. Calculate the surface integral ∫ ∫ zdydz, where ∑ is the lower side of the cone z = x ^ 2 + y ^ 2 between the plane z = 0 and z = 3
- 3. Find the volume of solid surrounded by x ^ 2 + y ^ 2 under the sign of cone z = root, x ^ 2 + y ^ 2 = 1 on cylinder and z = 0 on plane
- 4. This is an example in the book: finding the volume of the solid surrounded by a rotating paraboloid A: X & # 178; + Y & # 178; = R & # 178; and B: X & # 178; + Z & # 178; = R & # 178
- 5. ∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy, where the surface is outside the upper part of x ^ 2 + y ^ 2 + Z ^ 2 = 1
- 6. We know that x, y and Z are equal difference sequence. We prove that x2 [y + Z], Y2 [x + Z], Z2 [x + y] are equal difference sequence
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- 18. It is proved that the volume bounded by the tangent plane of any point on the cubic (a > o) of the surface XYZ = A and the three coordinate planes is a certain number The answer is: surface XYZ = A & # 179; the normal direction of (x0, Y0, Z0) is {y0z0, z0x0, x0y0} The tangent plane is: y0z0 (x-x0) + z0x0 (y-y0) + x0y0 (z-z0) = 0 Its intercept on three coordinate axes are 3x0, 3y0 and 3z0 The volume of the tetrahedron bounded by the tangent plane and three coordinate planes is: 27x0y0z0 / 6 = 9A & # / 2 My question is, how do you calculate the intercept above!
- 19. The geometric meaning of double integral is volume, but double integral can produce negative number,
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