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Let l be the circle x2 + y2 = 4, and the root of the curve integral (x2 + Y2) DS is equal to?
Root sign of curve integral (x2 + Y2) ds = root sign of curve integral (4) ds = curve integral 2ds = 2 * curve integral DS = 2 * circumference of circle = 2 * (2 * π * 2) = 8 π
S is x ^ 2 + y ^ 2 + Z ^ 2 = 4, find ∫ (x ^ 2 + y ^ 2) ds
Using the method of display expression: D = {(x, y): x ^ 2 + y ^ 2
Calculate the surface integral ∫ ∫ Z ^ 3DS, where s is the part of the hemispherical z = √ (a ^ 2-x ^ 2-y ^ 2) inside the conical z = √ (x ^ 2 + y ^ 2)
This is because every point in the integral function is on the given surface, and the projection surface is also a surface. Not all the coordinates satisfy z = √ a ^ 2-x ^ 2-y ^ 2. In fact, as long as Z = 0 is not satisfied, right?
It is known that the point a (radical 3,0) q is a moving point on the circle M = (x + radical 3) ^ 2 + y ^ 2 = 16, and the middle perpendicular of the segment AQ intersects MQ at the point P
(M is the center of the circle), find the trajectory equation of the moving point P?
By drawing a graph (you can draw a graph yourself), we can find that MP + AP = MP + PQ = R, so the sum of the distances between P point and two fixed points m and a is constant (i.e. the radius of the circle is 4), so we can find the trajectory equation of P point according to the definition of ellipse
RELATED INFORMATIONS
- 1. Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is at p. when the Q point moves on the circle, we can find Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is p. when Q moves on the circle, the trajectory of P is obtained
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