Help with calculus please.

[mashadar]

Hi,

i'm new on the forum, and think it is awesome !

I did several problems today, and i would like to double check if the logic is good, or if i'm just lucky with numbers...

Thanks to anyone who's taking the time to read

#1
Let C be the positively oriented square with vertices (0,1),(1,1),(1,0),(0,0)
Use Green's theorem to evaluate the line integral =­> integral over C of 2y^2xdx+8x^2ydy

I set P=8x^2y and dP/dx=16xy
I set Q=2y^2x and dQ/dy=4xy

double int (x and y from 0 to 1) of dP/dx-dQ/dy ... = 3

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#2
Suppose C is any curve from (0,0,0) to (1,1,1), and F(x,y,z)=< 2z+5y, z+5x, y+2x >, compute int over C F.dr

so i set r to be r(t)=< t,t,t > dr/dt = <1,1,1>
so now f is parametrized to `F(t)=< 7t, 6t, 3t >
now we dot prod dr/dt and F = 16t
integral from 0 to 1 of 16tdt = 8

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#3
If C is the curve given by r(t)=< 1+sint, 1+2(sint)^2, 1+3(sint)^3 >, t varies from 0 to pi/2 and F is the radial vector F(x,y,z)=< x,y,z >, compute the work done on a particule moving along C.

so x=1+sint, y=1+2sin^2(t), z=1+3sin^3(t) ... we find dx/dt .. dz/dt

then we dot prod F and the new dr(t) from 0 to pi/2

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#4
I still haven't found that one, if someone could help:

x^(2/3)+y^(2/3)=1
Find a parametrization of the curve and use it to compute the area of the interior...

I tried setting x=cos^3(t) and y=sin^3(t) which would give us r=1. but i don't really know what to do then... or even why i'm setting these values for x and y

Thanks for reading

 

[tkhunny]

#4 That seems the right thing to do.

\(\displaystyle x(t) = \cos^{3}(t)\)

\(\displaystyle y(t) = \sin^{3}(t)\)

For \(\displaystyle 0 \le t \le 2\pi\)

Substituting, \(\displaystyle x^{2/3} + y^{2/3} = \cos^{2}(t) + \sin^{2}(t) = 1\) for all points on the curve.

The next part is kind of magic. You need functions P(x,y) and Q(x,y) such that \(\displaystyle Q_{x}-P_{y}=1\). Where that "1" is the same "1" we just invented from our substitution in the original expression. You can just make up P and Q, but there is no advantage in making it more difficult than it needs to be.

How about P(x,y) = -y/2 and Q(x,y) = x/2? You can verify that \(\displaystyle Q_{x}-P_{y}=1\).

We're close!!

Using the standard notation:

\(\displaystyle F(x,y) = -\frac{y}{2}i + \frac{x}{2}j\)

r(t) is defined above: \(\displaystyle r(t) = <x(t),y(t)> = <\cos^{3}(t),\sin^{3}(t)>\)

\(\displaystyle F(r(t)) = <-\frac{\sin^{3}(t)}{2},\frac{\cos^{3}(t)}{2}>\)

\(\displaystyle r'(t) = <-3\cos^{2}(t)\sin(t),3\sin^{2}(t)\cos(t)>\)

FINALLY!!!!!

\(\displaystyle \int_{0}^{2\pi}\frac{3}{2}\cos^{2}(t)\sin^{4}(t) + \frac{3}{2}\sin^{2}(t)\cos^{4}(t)\;dt\;=\;\frac{3}{8}\int_{0}^{2\pi}\sin^{2}(2t)\;dt\;=\;\frac{3\pi}{8}\)

Wasn't that fun?!

In this case, you can verify directly, using only the first quadrant and exploiting symmetries:

\(\displaystyle 4\cdot\int_{0}^{1}\int_{0}^{\left[1-x^{2/3}\right]^{3/2}}\;dy\;dx\)

Which, if you are not hardy enough to evaluate that directly, using the substitution x = u^3 and y = v^3, gives the much simpler

\(\displaystyle 4\cdot\int_{0}^{1}\int_{0}^{\sqrt{1-u^{2}}}9u^{2}v^{2}\;dv\;du\)

In any case, it's all \(\displaystyle \frac{3\pi}{8}\)

Well, that is a lot to soak in. Thanks for making me remember how to do that. :shock:

 

[mashadar]

wow, it does look at first glance like magic

i'll take the time to read it and understand it!

Thank you

 

[mashadar]

The next part is kind of magic. You need functions P(x,y) and Q(x,y) such that \(\displaystyle Q_{x}-P_{y}=1\). Where that "1" is the same "1" we just invented from our substitution in the original expression. You can just make up P and Q, but there is no advantage in making it more difficult than it needs to be.

i get everything but why we need to find Qx-Py=1 ??

 

[tkhunny]

It's Green's Theorem, backwards. After we parameterized the boundary and substituted, it left us with "1" in the integral. Then we build what is usually the right-hand side of Green's Theorem. I tried to suggest this with "Where that "1" is the same "1" we just invented from our substitution in the original expression."

In other words, we turned the integrand into "1" with the substitution. To apply Green's Theorem, we need to make it look like Green's Theorem. Rather than some constant, we need some expression equivalent to 1.