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Answer:
The Line integral is π/2.
Step-by-step explanation:
We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:
we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence
(Remember that c is treated like a constant just for the x-variable).
This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.
Where, again, the constant of integration depends on Z.
As a result,
if we derivate f over z, we obtain
That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)
The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.
Just did a specific one of these; let's do the general case.
The point nearest the origin is (a,b).
The line from the origin through the point is
The line we seek is perpendicular to this one. We swap the coefficients on x and y, negating one, to get the perpendicular family of lines. We set the constant by plugging in the point (a,b):
That's standard form; let's plug in the numbers: