Curve Integral Solver Expert to Curve Integral Solver

The double integration part of the ydxdy integration region is the region enclosed by the l and x-axis, which can be expressed as.

x,y): x=s(t-sint) y=s(1-cost),0<=t<=2π ,0=

To find the amount of mass dissipation, the centroid can be found by using the formula of the trapped bridge difference centroid. The formula is shown in the figure below.

The answer is that Gao Lusha Zhen is excited as follows: Qizhou.

Let x=2cost,y=2+2sint from x 2+y 2=4y, and substitute the missing family into x 2+y 2+z 2=6y, and get z 2=2y=4+4sint,z=2 (1+sint).

dx=-2sintdt,dy=2costdt,dz=costdt/√(1+sint)

x^2+y^2-z^2=4y,y^2+z^2-x^2=6y-2x^2,z^2+x^2-y^2=6y-2y^2

Original selling loss = <0,2 >(4+4sint)(-2sint)dt+[6(2+2sint)-2(2cost) 2]*2costdt+[6(2+2sint)-2(2+2sint) 2]*costdt (1+sint).

4 Medium <0,2 >[2sint-2sin t+(6+6sint-4cos t)cost]dt+[3 (1+sint)-2(1+sint) (3 2))]d(1+sint)].

4∫<0,2π>[2sint-1+cos2t)dt+[2+6sint+4sin^t)]d(sint)+[3√(1+sint)-2(1+sint)^(3/2))]d(1+sint)

Substituting the expressions of x and y into the difference of the trapped part of the integrand, after a simple calculation, we know that the integral is commemorated = 0, so the integral is zero. Ruler board.

Select (a).

Sorry, no, I also read the prompt answer in the exercise book, the easy way is not known for the time being, but if it is a multiple-choice question, you can consider the elimination method The focus is to find the parametric equation of the curve, and then set the formula to calculate.

First, by subtracting z from the given two equations, we get x 2 + xy + y 2 = (a 2) 2

In the second step, let x=x1-y1 , y=x1+y1, get 3(x1) 2 + y1) 2=(a 2) 2

From this, the parametric equation for the curve is shown in Fig.

You should be able to calculate the rest of the formulas, so I won't calculate it in detail.

∫(e^xsiny+x+y)dx+(e^xcosy)dy

Now let's make up a straight line from (1,0) to (-1,0). So the semicircular arc from (-1,0) to (1,0) and the straight line from (1,0) to (-1,0) form a loop.

According to Green's formula, the integral of the curve on this loop can be reduced to a double area integral. Because the loop is not positive, there is an extra negative sign in front. Apply Green's formula.

∫e^xcosy-e^xcosy-1)dxdy

dxdy (i.e. semicircular area) = 2

Since a straight line from (1,0) to (-1,0) is calculated, the integrals on the straight line from (1,0) to (-1,0) are subtracted at the end, that is, the integrals on the straight line from (-1,0) to (1,0) are added.

The integration on this (-1,0) to (1,0) line is easy, because y is always a constant 0, dy is 0, and x is -1 to 1. Thereupon.

e^xsiny+x+y)dx+(e^xcosy)dy

xdx=½x²=0

So the end result is 2+0.

Using Green's formula, we know that the integral is independent of the path.

Then the path composed of two polyline segments, (1,0), (6,0), (6,8), is selected for integration.

Get (6-1) + (10-6) = 9 to choose A

As shown in the figure below, the basic formula for the surface division of the first type is used.

Make a straight line y=0 to make the curve into a closed curve (a semicircle), take the counterclockwise as the direction of the curve, and use Green's formula.

Original = (e x)cosy-[(e x)cosy-1]dxdy - 0->a] (e x)siny-y dx

The double integration interval d is (x-a 2) 2+y 2<=(a 2) 2,x>=0,y>=0

Original = dxdy- [0->a] (e x)sin0-0 dx = dxdy=(1 2) (a 2) 2=( a 2) 8

It can be found using the method in the complex variable function