Evaluate the line integral, where c is the given curve. ∫c sin(x) dx + cos(y) dy, where c consists of the top half of the circle x^2 + y^2 = 4 from (2, 0) to (−2, 0) and the line segment from (−2, 0) to (−3, 2)

Step-by-step explanation:

Previous concepts

Fundamental Theorem of Line Integrals and into Green’s Theorem

Suppose a Curve C given by the vector function with a= r(a) and b= r(b). Then we have this:

Where Vf represnent the gradient of the function f.

Solution to the problem

We want to find this integral:

And the region C is given by the top half circle with equation with radius 2 and we have a line segment from (-2,0) to (-3,2).

In order to solve this problem we can use this form:

Where is a bivariate function.

For our case our function F is given by: and that represent our vector field.

We can find the potential function or the antiderivate like this:

At the point (2,0) we can find the potential function and we got:

At the point (-3,2) we can find the potential function and we got:

And then the integral is given by:

what’s the question?

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.