Practical lesson. Calculating curvilinear integrals 1 and 2 kind
A practical lesson is devoted to calculating curvilinear integrals. Let's get started.
Task 1: we need to calculate the curvilinear integral (see the video). Note that there is a dl multiplier. This means that it is a curvilinear integral of the 1st kind, and we need dl to calculate it. Let's consider how the curve is set. It is set from point (0; 0) to point (2; 2). We need to express y in Cartesian coordinates. We can write with a plus sign or with a minus sign, but y in this section has a plus sign, so we write the root of 2x. To calculate dl, we use the formula: 1 plus the derivative of the function y by x squared, multiplied by dx (see the video).
First, we are going to consider how the value of the derivative is calculated. We multiply y’ (it is a constant factor, root of 2) by the derivative of the root of x (1 divided by 2 roots of x). We reduce it to this form (see the video).
We return to dl. This is the root of the equation: (see the video). We can proceed to calculate the integral. When we calculate a curvilinear integral, we have a curve. On a curve, the y value is calculated using the formula: the root of 2, instead of dl, we write what we get 2*x plus 1 divided by 2x plus 1 multiplied by dx. We can see that x changes from 0 to 2. Let us transform the equation. To calculate this integral, we write 2x plus 1 to the power of 1/2. We add 2x plus 1 with a coefficient of 1/2 under the sign of the differential; x changes from 0 to 2. We use the table integral of the power function, 1/2*(2x + 1)3/2 is multiplied by 2/3. The limits of integration are from 0 to 2. We reduce 2s and get the coefficient 1/3. Thus, 1/3*(53/2-1) = (5√5 – 1)/3. This is the answer.
We are going to move on to Task 2. ∫x2dl. Let's set a curve, it is marked here (this is part of a circle of radius a with the center at point O, located in the first quarter). We set it parametrically: x=a cos t, y = a sin t, where t is from 0 to π/2.
If the curve is set parametrically, the value of dl is calculated using the formula: the partial derivative of x squared plus the derivative of y squared, all multiplied by t. The derivative of the cosine is minus the sine, but it is squared, it is а2sin2t. The derivative y is accost; squaring it, we get а2cos2t; everything is under the root. We notice the trigonometric unit and get adt.
dl is calculated. Now we are going to calculate the integral along the curve L. In this case, x is calculated using the formula: а2cos2t, where t is from 0 to π/2, and we have adt (this is the dl multiplier). Next, we take а3 out of the integral sign and calculate it using the standard method. We apply the formula for reducing the degree to the cosine squared; this is 1 + cos 2t divided by 2, dt, the integration limits are from 0 to π/2. а3/2 is a multiplier. The integral of 1 is t, the integral of cos 2t is 1/2 sin 2t, the integration limits are from 0 to π/2. Note that the second term gives us 0 at both the upper and lower points, so it is sufficient to find t at the point π/2. Finally, we have πа3/4. The problem is solved.
Task 3. We come across a new notation. This circle on the integral sign (see the video) says that integration is carried out along a closed contour in the positive direction. The curve is closed, and L is the contour of a square. If the border traversal direction is positive, the domain should be on the left. Here, the positive direction is counterclockwise. There are 4 sections here, so we use the properties of the integral: this is the integral OC plus the integral CB plus the integral BA plus the integral AO. In contrast to the first kind of integral (this is already the second kind, because of the variable y), the integral depends on the direction of integration, so when writing, we follow OC, CB, BA, AO. We need to calculate each of these terms. The OC integral: in the section OC y = 0, it means that dy (the differential of the constant) is zero. The CB integral: we set the integration section CB, here x = 1, and y changes from 0 to 1. We get a definite integral. Calculating it, we see that y is from 0 to 1 plus 1/2 y squared from 0 to 1. 1 plus 1/2 = 3/2 is the second term. The third section: the integral of the BA curve. Here y = 1, the differential is a constant (equal to zero), so we can immediately write that the third term is 0. The last section: the integral along the AO curve. Here x = 0, and y changes from 1 to 0. The integration limits are as follows: the lower one is 1 and the upper one is 0. We get 1/2 y squared from 1 to 0. 0 minus 1/2 equals minus 1/2. We can proceed to calculating the answer. Two integrals are equal to zero, one is equal to 3/2 and the last is equal to -1/2. The answer is 1. We have solved the problem.