Vector Calculus Help! Please!!!?

Question

Let vector F(x,y)=<√y , √x> be a force field. Let R be the region bounded by y=0, x=2 and y= (x^3)/4. Let C be the closed curve, oriented counter-clockwise, which is the boundary of region R.

A. Draw a sketch of R and C. (I know...you can't do it on here...just giving all the instructions).
B. Using a Line Integral, compute the work done by the force field F on an object that travels one time around the path C.
C. Compute the following double integral:
∫ ∫½((x^-½) - (y^-½))dA
D. Does Green's Theorem say that your answer from part b and c should be equal? Are they equal?

Okay...so I know that I can't use Green's Theorem on the region enclosed by curve C, since F does not have first order partial derivatives at the origin. But the way that they work the problem out makes no sense to me....So please could a GENIUS help me with the problem...Please explain it so that I can understand!!!! Thanks so much for your work! (And your brain power! haha!) =)

Answer 1

Find the work done on each section of the path separately. Split it up into three parts where each part is along a curve that can be represented by a single equation. In your case, one path will be along the line y = 0 from x=0 to x=2.

You will want to use the following formula to find the work
b
∫ F(c(t))•c'(t)dt
a

F(c(t)) gives the force at each point along the path.
c'(t) gives the velocity of a point moving along the path.
dt a infinitesimal amount of time.
Since distance is equal to velocity times time, this formula makes sense. c'(t)dt is just velocity times time = distance.
So you integrate along the path to sum up all the (Force times distance) pieces which is what work is.

This formula requires a parametrization of the path. It doesn't matter what parametrization you use. It will still come out the same. Just make it simple.
For the first section I noted above this would could be parametrized by c(t) = (t , 0) from t=0 to t=2.

If you understand work well though you can see that you don't need to do an integral for this part. The force field is at a right angle to the direction of movement along this path (the x-component of the force field is zero). This means that the work is zero.

You just need to figure out how to parametrize the other two parts then evaluate the integrals.

Good luck!

Answer 2

Calculus 2 has some "in intensity" calculus like numerous integration recommendations etc., as others have remarked. Vector calculus incorporates the thank you to handle and do calculus with vectors. this is straight forward easily. you have some concepts like Gauss's theorem, green's theorem and a few ideas like divergence, curl etc., that are surprisingly hardship-free. it variety of feels greater suited to me in case you're taking the vector calculus with purposes direction. you do no longer want Calculus 2 for doing vector calculus effectively. common calculus is adequate. yet once you're apprehensive which you will get at a loss for words in case you circulate away maths for one total semester then you definately can take calc 2. yet as a pronounced earlier Vector Calculus is greater uncomplicated and you do no longer want calc 2 for vector calculus.

Answer 3

Maybe try this:

(Line) integral Fdx+ Fdy= (Green) integral (dF/dy) dy. dx + (dF/dx) dx dy
= integral dF/dx . dx*dy - dF/dy dx*dy

or you can do affine transformation to u, v coordinates(another 3 vertex) on the u,v that doesn't include 0 and do a pull back.