Using polar coordinates to calculate double integral ∫ [D] (6-3x-2y) DXDY =? Where D: x ^ 2 + y ^ 2

Using polar coordinates to calculate double integral ∫ [D] (6-3x-2y) DXDY =? Where D: x ^ 2 + y ^ 2

Let x = RCOs θ, y = rsin θ, then 0 < R < R, 0 < θ < 2 π. So the original integral = ∫ (0 to 2 π) d θ∫ (0 to R) (6-3rcos θ - 2rsin θ) RDR
=∫ (0 to 2 π) [(3R ^ 2-r ^ 3cos θ - 2 / 3 × R ^ 3sin θ) (r = R) - (3R ^ 2-r ^ 3cos θ - 2 / 3 × R ^ 3sin θ) (r = 0)] d θ
=R ^ 2 ∫ (0 to 2 π) [(3-rcos θ - 2 / 3 × rsin θ) d θ
=R^2×(3θ-Rsinθ+2/3×Rcosθ)(θ=2π)-R^2×(3θ-Rsinθ+2/3×Rcosθ)(θ=0)
=6πR^2.

∫∫∫ x ^ 3 + 3x ^ 2Y + 3xy ^ 2 + y ^ 3 DXDY integral region is symmetric about X axis, why can the original formula be simplified to ∫∫ x ^ 3 + 3xy ^ 2 DXDY
∫∫ x ^ 3 + 3x ^ 2Y + 3xy ^ 2 + y ^ 3 DXDY integral region is symmetric about X-axis
∫∫ x ^ 3 + 3xy ^ 2 DXDY this question is the 16th question of 2010 postgraduate entrance examination mathematics (3)

3x ^ 2Y + y ^ 3 is an odd function of Y
If the integral region is symmetric about X axis, then its integral is 0

Solve the equation of vertical straight line passing through point (1,2, - 3) and plane 3x-2y + 2Z + 3 = 0

If the normal vector of plane 3x-2y + 2Z + 3 = 0 is {3, - 2,2}, the linear equation passing through point (1,2, - 3) is as follows:
x-1/3=y-2/-2=z+3/2

The linear equation passing through (- 1,3,2) and perpendicular to the plane 3x-2y + Z + 5 = 0 is

The linear equation passing through (- 1,3,2) and perpendicular to the plane 3x-2y + Z + 5 = 0 is
（x＋1）/3=（y-3）/（-2）=（z-2）/1

Find the plane which passes through point (1,1,1) and is perpendicular to two planes X-Y + Z = 4,3x + 2y-12z + 5 = 0

Let the plane vector of the plane be m = (a, B, c), then we know that the plane vectors of the two planes are a = (1, - 1,1), B = (3,2, - 12)
So we combine the equation A-B + C = 0 with the equation 3A + 2b-12c = 0, eliminate the element B, let C = 1, get a = 2, B = 3, then let the plane equation be 2x + 3Y + Z = m, and substitute the point (1,1,1) to get m = 6, so the equation is 2x + 3Y + Z = 6,

Finding the projection of the line lx-2y-z + 7 = 0, 2Y + 3z-5 = 0 on the plane X-Y + 3Z + 8 = 0

By x-2y-z + 7 = 0, 2Y + 3z-5 = 0, the direction vector of the straight line can be obtained, which is the vector N1 perpendicular to the normal vectors of the two planes. By changing the vector and the normal vector N2 of the plane X-Y + 3Z + 8 = 0, another normal vector N3 (a, B, c) can be obtained

Find the equation of plane π passing through the line l3x-2y + 2 = 0, x-2y-z + 6 = 0 and the distance from the point M0 (1,2,1) is 1

The expression of this problem is not very clear. L should be the equation of intersection, that is, the intersection of plane 3x-2y + 2 = 0, x-2y-z + 6 = 0
Let x-2y-z + 6 + K (3x-2y + 2) = 0,
That is, (1 + 3K) X-2 (1 + k) Y-Z + 6 + 2K = 0,
The distance from point M0 (1,2,1) to the desired plane
d=|(1+3k)-4(1+k)-1+6+2k|/ √[(1+3k)^2+4(1+k)^2+1]
=|2+k|/ √(6+14k+13k^2)=1,
That is, 4 + 4K + K ^ 2 = 6 + 14K + 13K ^ 2, then 6K ^ 2 + 5K + 1 = 0, k = - 1 / 3 or K = - 1 / 2
The equation of plane π is 4Y + 3z-16 = 0, or x + 2Y + 2z-10 = 0

Solve the plane equation which passes through the point (1,1,1) and is perpendicular to the plane X-Y + z-7 = 0 and 3x + 2y-12z + 5 = 0 at the same time

The equation of the intersection of two planes is the normal of the plane to be solved. List the normal vector and use the point method to solve it. The method to solve the direction vector of the intersection of two planes (that is, the normal vector of the plane to be solved) is: using the determinant, we can get the following formula: I = 12-2, j = 3 + 12, k = 2 + 3, the normal vector of the plane to be solved is {I, J, K,} that is {10, 15, 5}, list the points

The projection equation of the line X-1 = y = 1-z on the plane X-Y + 2z-1 = 0 is

Line (x-1) / 1 = Y / 1 = (Z-1) / (- 1) s = (1,1, - 1) plane X-Y + 2z-1 = 0n0 = (1, - 1,2) let plane m pass through line X-1 = y = 1-z and plane X-Y + 2z-1 = 0 perpendicular I j kn = 1,1 - 1 = 2i-j-k-k-2j-i = (1, - 3, - 2) 1 - 1 2 plane M: x-3y-2z + D = 0 plane m passes through point (1,0,1) 1-2 + D = 1 plane M: x-3

Find the equation of line L1: (x-1) / 1 = (y + 1) / 2 = Z / 3 projecting line L on plane π, x + y + 2z-5 = 0

If the projection of line L1 on plane π is required, only the set of points constituting line L1 and the set of projection points on plane π need to be known
From the equation of line L1, it is easy to know that the set of points on line L1 is
From the set meaning of plane, we can see that the points on plane π are the set of points whose inner product is 5 with vector (1,1,2), so the normal vector of plane π is (1,1,2)
Thus, the line between point (T + 1 + X, 2t-1 + X, 3T + 2x) and point (T + 1,2t-1,3t) is perpendicular to plane π (because it is parallel to the normal vector of π)
 
Let the point (T + 1 + X, 2t-1 + X, 3T + 2x) be on the plane π, that is, the set of projections of L1 on the plane π is, then (T + 1 + x) + (2t-1 + x) + 2 (3T + 2x) - 5 = 0, and the solution is obtained
 
So the set of L1 projection points is
The equation of projection line L is obtained