Question

more help with green's theorem

Answer 1

alright im up to this point and Im not sure what to put in for the bounds. I know I use the op surface and it would be an ellipse but I am not sure how to find that equation and what the bounds would be. or can I use the bottom surface and use the circle equation given?

Answer 2

@IrishBoy123

Answer 3

@freckles

Answer 4

@Vocaloid

Answer 5

@imqwerty

Answer 6

@KendrickLamar2014

Answer 7

@Compassionate

Answer 8

this may or may not help, ....my scribblings

in terms of limits you are projecting the ellipse onto the circle....so i think it simplifies considerably.

Answer 9

@jagr2713

Answer 10

im still lost im not sure what you're doing with da

Answer 11

why is it secgama?

Answer 12

can I solve for the positive root of y in the circle equation plug that in for the plane and use that as the projection?

Answer 13

you are projecting the area onto the xy plane.

so you can say that \(da \cos \gamma = dx \, dy\) or \(da = \dfrac{dx \, dy}{\cos \gamma} = \sec \gamma \, dx \, dy\)


https://i.gyazo.com/c4c40770961a72b24e5646536e73abcb.png

Answer 14

that's different from what we were shown for some reason our n doesn't have to be a unit vector as well so I don't have the sqrt(2)

Answer 15

let me scribble some stuff out, the latex is great but really time consuming....

Answer 16

I'll just get help from a tutor on monday

Answer 17

personal tutor is great way to operate, good luck

Answer 18

you found the curl(F)= (0,-1,3)
that is ok
the plane z= 1+y

we want to compute:
\[ \int \int_S curl(\vec{F}) \cdot \hat{n} \ dS \]
given z= f(x,y) , we can find
\[ \hat{n} \ dS = <-f_x, -f_y, 1> dx\ dy \](note that the vector is not \( \hat{n} \) and \(dx \ dy\) is not dS
see
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-28-divergence-theorem/

with f(x,y) = 1+y (i.e. this is the plane)
<-fx, -fy, 1> is <0, -1, 1>
and the integral becomes
\[ \int \int_S curl(\vec{F}) \cdot \hat{n} \ dS \\
\int \int_S <0,-1,3>\cdot <0,-1,1> \ dx \ dy \\
4 \int \int_S \ dx \ dy
\]

The bounds are the "shadow" of the surface S projected onto the x-y plane. In this case, the shadow is the unit circle, and we know the area will be \(\pi\). Thus the answer is
\[ 4 \pi\]

Answer 19

We could also do it this way:
The equation of the plane in standard form is 0x -1y + 1z = 1
or (0,-1,1) dot P = 1
and the unit normal is \[ \hat{n}= \frac{<0,-1,1>}{\sqrt{2}} \]
\[ \int \int_S curl(\vec{F}) \cdot \hat{n} \ dS \\
\int \int_S <0,-1,3> \cdot <0, -1,1>/\sqrt{2} \ dS \\
2\sqrt{2} \int \int_S \ dS \]

from z= 1+y where y varies from -1 to 1, we find the semi-major axis "a" of the ellipse is \( \sqrt{2}\) and the semi-minor axis "b" is 1. The area of an ellipse is \(a b \pi\)
thus the area of the ellipse is \( \sqrt{2} \pi\)
\[ \int \int_S \ dS = \sqrt{2} \pi \]
and the answer is
\[ 2\sqrt{2} \cdot \sqrt{2} \pi= 4 \pi \]