![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
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
If the surface is outside the surface of the solid surrounded by the cone z = radical (x ^ 2 + y ^ 2) and z = 1, then ∫ ∫ xdydz + ydzdx + zdxdy=
Gauss formula can be used directly
If there's no problem, please take it, / > if there's no problem
Make the tangent of circle x ^ 2 + y ^ 2 + 2x-4y = 0 through point a (4,2), and find the tangent equation
X ^ 2 + y ^ 2 + 2x-4y = x ^ 2 + 2x + 1 + y ^ 2-4y + 4 = 5 (x + 1) ^ 2 + (Y-2) ^ 2 = 5 when the tangent slope does not exist, it is obvious that the tangent equation does not hold. Let Y-2 = K (x-4) kx-y-4k + 2 = the distance from the point 0 to the straight line | - k-2-4k + 2 | / √ K ^ 2 + 1 = √ 525K ^ 2 = 5K ^ 2 + 520k ^ 2 = 5K = ± 1 / 2, so x-2y = 0-x-2y + 2 + 2 = 0x + 2y-4
Through a point outside the circle (3,5), make a circle: x ^ 2 + y ^ 2-2x + 4Y + 1 = 0 tangent, find tangent equation
It's very simple~
Center coordinates: (1, - 2) radius r = 2
It is found that x = 3 is just a tangent of it
To find another tangent, first set the oblique formula: Y-5 = K (x-3)
Then the distance from the center of the circle to the straight line is radius 2, and K = 45 / 28 can be obtained
So the two tangent equations are solved
There must be two tangents to make a circle through a point outside the garden!
The equation of tangent passing through point P (2,3) leading circle x ^ 2 + y ^ 2-2x + 4Y + 4 = 0 is
Do with the point oblique formula, write the calculation process of the root formula in detail
First find out the center and radius (x-1) ^ 2 + (x + 2) ^ 2 = 1
The radius of the center (1, - 2) is 1
There are two equations through this point
Draw your own picture, you can see that x = 2 is one of the tangent equations
Let all other points be (a, b)
List two equations into the point
（2-a）^2+（3-b）^2=25
（1-a）^2+（-2-b）^2=1
We get a = - 8-5b
Then the point (- 8-5b, b) is substituted into the original equation
We get b = - 2 or 21 / 13
Two tangent points (2, - 2) and (1 / 13,21 / 13) were determined
It is easy to get a tangent line x = 2 and 25y-18x-39 = 0
Through the point P (- 2, - 3), make the tangent of the circle x ^ 2 + y ^ 2-2x-4y-4 = 0, and find the tangent equation
（x-1）^2 + (y-2) ^2 =9
(1,2) center
Point a (- 2,2) is on the circle,
So p (- 2, - 3) and a are in a straight line,
It's perpendicular to the x-axis
The tangent is x = - 2
Through the point P (- 2, - 3), make the tangent line of the circle x ^ 2 + y ^ 2-2x -- 4y-4 = 0, and find the equation of the tangent line
If P (- 2, - 3) is substituted into a circle, we can know that P (- 2, - 3) should have two tangents outside the circle. Let the linear equation of P (- 2, - 3) be x + by + 2 + 3B = 0, and directly substitute it into the equation of the circle to get y & # 178; + (- by-2-3b) &# 178; - 4Y-2 (- by-2-3b) - 4 = 0, and simplify to (B & # 178; + 1) y & # 178; + (6B & # 178; + 6b-4) y + (9b & # 17
The graph enclosed by y = x ^ 2 and y = 2-x ^ 2
Seek the area of plane figure, not drawing!
To solve the simultaneous equation: y = x & sup2;, y = 2 - X & sup2; 2x & sup2; = 2, x = ± 1 ∵ y = 2 - X & sup2;, the area enclosed by the image above y = x & sup2; is = ∫ [(2 - X & sup2;) - X & sup2;] DX (X: - 1 → 1) = ∫ (2 - 2x & sup2;) DX (X: - 1) = ∫ (2 - 2x & sup2;)
In double integral, if the integral region is symmetric with respect to y = x, the integrand f (x, y) = f (y, x)
When the integral region of the double integral is symmetric with respect to y = x, is the integrand function x and y of the double integral of y = x the same if they are interchanged?
Is that to say, like "definite integral has nothing to do with integral variable", double integral has nothing to do with integral variable?
X, y switching will not affect the result, but will affect the calculation speed, so it is important to choose the integral variable
When the integral domain D is symmetric with respect to the straight line y = x, the two variables of the integrand in the double integral can exchange positions,
How did this property come about?
If D is symmetric with respect to the line y = x, then
therefore
The absolute value of that determinant is Jacobian matrix. I think you have learned the binary integral substitution method
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. ∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy, where the surface is outside the upper part of x ^ 2 + y ^ 2 + Z ^ 2 = 1
- 4. 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
- 5. X2 + Y2 + Z2 = 1 to find the value range of X + y
- 6. Find the intersection of the plane a: 3x-4y Z 7 = 0 and the line L: x-3y + 12 = 0, 2y-z-6 = 0 The equation of a line perpendicular to a line in a given plane
- 7. How to solve the quadratic equation of one variable 3x * 2-5x + 2 = 0, ;
- 8. How to solve the equation of 2.8 × 4 + 5x = 15.8
- 9. The linear equation of the intersection line passing through the point (- 3,2,5) and parallel to the plane x-4z = 0 and 2x-y-5z = 1
- 10. Find the plane parallel to the line connecting point a (2,5, - 3) and point B (3, - 2,2) through the intersection of two planes π 1:2x + Y-Z + 1 = 0 and π 2: x + y + 2Z + 1 = 0 equation
- 11. Find the volume of solid surrounded by z = x ^ 2 + y ^ 2 and z = 2-radical (x ^ 2 + y ^ 2)
- 12. 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?
- 13. Normal equation of curve y = x ^ 3-3x at point (2,2) If the problem, solve!
- 14. If the center of the ellipse is the origin, the focus is on the x-axis, the point P is a point on the ellipse, and the projection of P on the x-axis is exactly the origin of the ellipse The center of an ellipse is at the origin, the focus is on the x-axis, P is a point on the ellipse, the projection of P on the x-axis is just the left focus of the ellipse, the line between P and center O is parallel to the line between the right vertex and the upper vertex, and the distance between the left focus and the left vertex is equal to √ 10 - √ 5. Try to find the eccentricity and equation of the ellipse
- 15. How to calculate the triple integral of a volume bounded by a rotating paraboloid and a plane perpendicular to the Z-axis in spherical coordinates
- 16. Find the intersection angle of two curves x + y = 0, XY '= 0 on the surface z = XY The two curves should be x + y = 0, xy = 0
- 17. Calculate ∫ ∫ ∑ (x ^ 2 + y ^ 2) ds, where ∑ is the part of paraboloid z = 2-x ^ 2 + y ^ 2 above the xoy plane If ∑ is the entire boundary surface bounded by the cone z = √ (x ^ 2 + y ^ 2) and the plane z = 1, how to find it?
- 18. Surface x2 + Y2 + Z = 9 tangent plane equation at point 124
- 19. The tangent plane equation at point P (1.1.1)
- 20. 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!