Question

Is this an example of a line integral question?

Answer 1

@triciaal

Answer 2

I have a quiz over line integrals tomorrow morning and need some things to practice.

Answer 3

@mathmale

Answer 4

I dont think this is actually a line integral question. I think its a double integral.

Answer 5

This may be a better example:

Answer 6

For the first question I got 5/2 as my answer

Answer 7

@Luigi0210 @sleepyjess

Answer 8

Yeah that first question is definitely a line integral

Answer 9

\[H( r(u) ) = -u^2i+2u^3j\]\[r'(u)=6u^2i-2uj\]
\[eq1*eq2=-10u^4\]\[\int -10u^4=-2u^5\]from 0 to 1 = -2

Answer 10

\(2u^3 \) should correspond to the i value @stamp

Answer 11

I mix things up all the time lol good eye

Answer 12

\[(2u^3, -u^2) \cdot (6u^2, -2u) \rightarrow \int\limits_{0}^{1}12u^3u^5+2u^3du \]
\[\int\limits_{0}^{1}12u^5+\int\limits_{0}^{1}2u^3=[2u^6+\frac{ 1 }{ 2 }u^4]_{0}^{1}=\frac{ 5 }{ 2 }\]

Answer 13

Me too @stamp

Answer 14

Conceptually you have to imagine h(x,y) as being some vector field in space. Like at every point (x,y) you have some flow of water. Now imagine you're a fish, swimming along the line r(u). r is the literal point you're at, while u is the parameter that tells you the progress [0,1] from 0 to 100% complete on your path. Now if you take this path, how much force of water might you feel from the water? Well that all depends on how much you're going with or against the flow at every point along your path, and that naturally can be interpreted as the dot product.

So at one point along your path, \(\vec r\) you'll have a little velocity vector there, \(\frac{d \vec r}{du}\) and the dot product with this with the flow of water you're swimming against is \(\vec h\). So like we described, the dot product will be a measure of that strength you need to swim at that point,

\[\vec h \cdot \frac{d\vec r}{du}\]

But this is just a tiny bit during a little point on our path, we need to add up all the contributions over all time:

\[\int_0^1 \vec h \cdot \frac{d\vec r}{du}du\]
Since we only care about the \(\vec h\) at the point we're at, \(\vec r\) we use the same coordinates there,

\[\int_0^1 \vec h(x(u),y(u)) \cdot \frac{d\vec r}{du}du\]

So now you can plug everything in.

Answer 15

One sec they are kicking me out of starbucks

Answer 16

\(\color{blue}{\text{Originally Posted by}}\) @legomyego180
\(2u^3 \) should correspond to the i value @stamp
\(\color{blue}{\text{End of Quote}}\)
Nah you're wrong, @stamp had it right the first time!

You're plugging it in as if it was:
\[\vec h(x,y) = x \hat \imath + y \hat \jmath\]
But it's really:
\[\vec h(x,y) = y \hat \imath + x \hat \jmath\]

Answer 17

For the second question, to verify that something is the gradient of a scalar field, you just have to take its curl and see that it's zero:
\[ \vec \nabla \times \vec h= 0\]
Since this is always true:
\[ \vec \nabla \times ( \vec \nabla f) = 0\]

From there, you gotta basically solve the differential equations for f so that you can evaluate it at the endpoints like they ask. Or if they hadn't asked you to solve it in this specific way, once you know that it's the gradient of a scalar field, you can take any path you want between your two points, usually one of the easiest ways to do that is to keep all your coordinates constant except 1 and move along the edges of a rectangle to your point.

Answer 18

Oh yea, youre right. Sorry @stamp. Told you I get stuff mixed up too haha. Thank you for the explanation @Kainui that helps me visualize it. Im looking at this second question now...