The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

WD Line Integrals

Scroll to see replies

Reply 20
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
thanks looking at it now! I'll have another solid go and see if I get the same. I'll let you know by tomorrow.


just don't ask anything right now

Homer-asleep2-300x209.gif
Original post by cooldudeman
Calculate the work done by the force field F(x,y,z)=(y2,z2,x2)F(x,y,z)=(y^2,z^2,x^2) along the curve of intersection of the sphere x2+y2+z2=1x^2+y^2+z^2=1, the cylinder x2+y2=xx^2+y^2=x, and the halfspace z>0z>0. The path is traversed in a direction that appears clockwise when viewed from the high above the xyxy-plane.

How do I even find the path for starters... after that it is straight forward.

I know what the sphere would look like in the xyz plane but not the cylinder. How do you sketch the cylinder?


I've now done this and got the same result as TeeEm. Here's a sketch with many details omitted:

1. The path is a tear drop shaped curve, symmetrical in +ve and -ve y.
The path segments for +ve, -ve y are, in vector form:

r+=(x,x(1−x),1−x)\bold{r}_+ = (x, \sqrt{x(1-x)}, \sqrt{1-x})
r−=(x,−x(1−x),1−x)\bold{r}_- = (x, -\sqrt{x(1-x)}, \sqrt{1-x})

where x∈[0,1]x \in [0,1]

2. I parameterise this with x=sin⁡2θx=\sin^2\theta with θ∈[0,π/2]\theta \in [0,\pi/2] which is a bijection. This gives:

r+=(sin⁡2θ,sin⁡2θ2,cos⁡θ)⇒dr+=(sin⁡2θ dθ,cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_+ = (\sin^2\theta, \frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_+ = (\sin2\theta \ d\theta, \cos2\theta \ d\theta, -\sin\theta \ d\theta)

r−=(sin⁡2θ,−sin⁡2θ2,cos⁡θ)⇒dr−=(sin⁡2θ dθ,−cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_- = (\sin^2\theta, -\frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_- = (\sin2\theta \ d\theta, -\cos2\theta \ d\theta, -\sin\theta \ d\theta)

and

F(x,y,z)=(sin⁡22θ4,cos⁡2θ,sin⁡4θ)\bold{F}(x,y,z) = (\frac{\sin^2 2\theta}{4}, \cos^2\theta, \sin^4\theta)

3. I'll leave the rest to you; you have to split up the path, and integrate ∫F⋅dr\int \bold{F} \cdot d\bold{r} over both bits with the appropriate limits to take you clockwise around it when viewed from above. Some nice cancellation occurs.
Reply 22
Avatar for TeeEm
TeeEm
19
Original post by atsruser
I've now done this and got the same result as TeeEm. Here's a sketch with many details omitted:

1. The path is a tear drop shaped curve, symmetrical in +ve and -ve y.
The path segments for +ve, -ve y are, in vector form:

r+=(x,x(1−x),1−x)\bold{r}_+ = (x, \sqrt{x(1-x)}, \sqrt{1-x})
r−=(x,−x(1−x),1−x)\bold{r}_- = (x, -\sqrt{x(1-x)}, \sqrt{1-x})

where x∈[0,1]x \in [0,1]

2. I parameterise this with x=sin⁡2θx=\sin^2\theta with θ∈[0,π/2]\theta \in [0,\pi/2] which is a bijection. This gives:

r+=(sin⁡2θ,sin⁡2θ2,cos⁡θ)⇒dr+=(sin⁡2θ dθ,cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_+ = (\sin^2\theta, \frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_+ = (\sin2\theta \ d\theta, \cos2\theta \ d\theta, -\sin\theta \ d\theta)

r−=(sin⁡2θ,−sin⁡2θ2,cos⁡θ)⇒dr−=(sin⁡2θ dθ,−cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_- = (\sin^2\theta, -\frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_- = (\sin2\theta \ d\theta, -\cos2\theta \ d\theta, -\sin\theta \ d\theta)

and

F(x,y,z)=(sin⁡22θ4,cos⁡2θ,sin⁡4θ)\bold{F}(x,y,z) = (\frac{\sin^2 2\theta}{4}, \cos^2\theta, \sin^4\theta)

3. I'll leave the rest to you; you have to split up the path, and integrate ∫F⋅dr\int \bold{F} \cdot d\bold{r} over both bits with the appropriate limits to take you clockwise around it when viewed from above. Some nice cancellation occurs.


post 19 any good?
Reply 23
Avatar for TeeEm
TeeEm
19
Original post by atsruser
I've now done this and got the same result as TeeEm. Here's a sketch with many details omitted:

1. The path is a tear drop shaped curve, symmetrical in +ve and -ve y.
The path segments for +ve, -ve y are, in vector form:

r+=(x,x(1−x),1−x)\bold{r}_+ = (x, \sqrt{x(1-x)}, \sqrt{1-x})
r−=(x,−x(1−x),1−x)\bold{r}_- = (x, -\sqrt{x(1-x)}, \sqrt{1-x})

where x∈[0,1]x \in [0,1]

2. I parameterise this with x=sin⁡2θx=\sin^2\theta with θ∈[0,π/2]\theta \in [0,\pi/2] which is a bijection. This gives:

r+=(sin⁡2θ,sin⁡2θ2,cos⁡θ)⇒dr+=(sin⁡2θ dθ,cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_+ = (\sin^2\theta, \frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_+ = (\sin2\theta \ d\theta, \cos2\theta \ d\theta, -\sin\theta \ d\theta)

r−=(sin⁡2θ,−sin⁡2θ2,cos⁡θ)⇒dr−=(sin⁡2θ dθ,−cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_- = (\sin^2\theta, -\frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_- = (\sin2\theta \ d\theta, -\cos2\theta \ d\theta, -\sin\theta \ d\theta)

and

F(x,y,z)=(sin⁡22θ4,cos⁡2θ,sin⁡4θ)\bold{F}(x,y,z) = (\frac{\sin^2 2\theta}{4}, \cos^2\theta, \sin^4\theta)

3. I'll leave the rest to you; you have to split up the path, and integrate ∫F⋅dr\int \bold{F} \cdot d\bold{r} over both bits with the appropriate limits to take you clockwise around it when viewed from above. Some nice cancellation occurs.


ignore me

I just saw the initial comment
Original post by atsruser
I've now done this and got the same result as TeeEm. Here's a sketch with many details omitted:

1. The path is a tear drop shaped curve, symmetrical in +ve and -ve y.
The path segments for +ve, -ve y are, in vector form:

r+=(x,x(1−x),1−x)\bold{r}_+ = (x, \sqrt{x(1-x)}, \sqrt{1-x})
r−=(x,−x(1−x),1−x)\bold{r}_- = (x, -\sqrt{x(1-x)}, \sqrt{1-x})

where x∈[0,1]x \in [0,1]

2. I parameterise this with x=sin⁡2θx=\sin^2\theta with θ∈[0,π/2]\theta \in [0,\pi/2] which is a bijection. This gives:

r+=(sin⁡2θ,sin⁡2θ2,cos⁡θ)⇒dr+=(sin⁡2θ dθ,cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_+ = (\sin^2\theta, \frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_+ = (\sin2\theta \ d\theta, \cos2\theta \ d\theta, -\sin\theta \ d\theta)

r−=(sin⁡2θ,−sin⁡2θ2,cos⁡θ)⇒dr−=(sin⁡2θ dθ,−cos⁡2θ dθ,−sin⁡θ dθ)\bold{r}_- = (\sin^2\theta, -\frac{\sin 2\theta}{2}, \cos\theta) \Rightarrow d\bold{r}_- = (\sin2\theta \ d\theta, -\cos2\theta \ d\theta, -\sin\theta \ d\theta)

and

F(x,y,z)=(sin⁡22θ4,cos⁡2θ,sin⁡4θ)\bold{F}(x,y,z) = (\frac{\sin^2 2\theta}{4}, \cos^2\theta, \sin^4\theta)

3. I'll leave the rest to you; you have to split up the path, and integrate ∫F⋅dr\int \bold{F} \cdot d\bold{r} over both bits with the appropriate limits to take you clockwise around it when viewed from above. Some nice cancellation occurs.


Thanks so much.

Posted from TSR Mobile
Original post by TeeEm
ignore me

I just saw the initial comment


I'm just confused on this.

Posted from TSR Mobile
1422641867272.jpg33.8KB
Reply 26
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
I'm just confused on this.

Posted from TSR Mobile


not quite that

z>0 and 0<x<1
Original post by TeeEm
not quite that

z>0 and 0<x<1


Ok if thats the case then why isn't it just the top right quadrant??

Posted from TSR Mobile
Reply 28
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
Ok if thats the case then why isn't it just the top right quadrant??

Posted from TSR Mobile



it is octant 1 and octant 4
Original post by TeeEm
it is octant 1 and octant 4


Octant? That means like 1/8 of the arc of a circle right?

I dont understand what you mean...

z>0 so how can the bottom right quadrant be involved at all?

Posted from TSR Mobile
Reply 30
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
Octant? That means like 1/8 of the arc of a circle right?

I dont understand what you mean...

z>0 so how can the bottom right quadrant be involved at all?

Posted from TSR Mobile


octant is 1/8 of a sphere

what exactly bothers you because I do not follow the issue...
Original post by TeeEm
octant is 1/8 of a sphere

what exactly bothers you because I do not follow the issue...


Ok so the intersection of the two is given as that pic which u have and I copied down just now. We're finding the path of when it is (in the zx plane) (0,1) to (1,0) then (1,0) to (0,-1) right?

But we are given the condition that z>0 so that means everything in the bottom quadrants should be ignored but then why are we finding the path of (1,0) to (0,-1)?

Posted from TSR Mobile
Reply 32
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
Ok so the intersection of the two is given as that pic which u have and I copied down just now. We're finding the path of when it is (in the zx plane) (0,1) to (1,0) then (1,0) to (0,-1) right?

But we are given the condition that z>0 so that means everything in the bottom quadrants should be ignored but then why are we finding the path of (1,0) to (0,-1)?

Posted from TSR Mobile


I think you do not understand/interpret what I am showing

THE PROFILES THAT I AM SHOWING ARE NOT THE REQUIRED CURVE
IT IS MERELY TRANSFORMATIONS OF A PARABOLA IN ORDER TO FIND SUITABLE PARAMETRICS.
Original post by TeeEm
I think you do not understand/interpret what I am showing

THE PROFILES THAT I AM SHOWING ARE NOT THE REQUIRED CURVE
IT IS MERELY TRANSFORMATIONS OF A PARABOLA IN ORDER TO FIND SUITABLE PARAMETRICS.


I see so how did you come up with the limits? Hpw did you just know them.

Posted from TSR Mobile
Reply 34
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
I see so how did you come up with the limits? Hpw did you just know them.

Posted from TSR Mobile


I will show you in an isolated example in about an hour.
Original post by TeeEm
I will show you in an isolated example in about an hour.


OK thanks. I think I understand everything beaides how you found the limits of t.

Posted from TSR Mobile
Reply 36
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
OK thanks. I think I understand everything beaides how you found the limits of t.

Posted from TSR Mobile


see if this helps

IMG.pdf
Original post by TeeEm
see if this helps

IMG.pdf



Thanks so much for this. Feel free to ignore this if you're not bothered to explain anymore but I still don't understand how you got the limits of t.

How did you deduce that the limits for x is (0,1)?
Also is the z^2=1-x a plane or just a curve?
Reply 38
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman


How did you deduce that the limits for x is (0,1)?


look at any of the 2D plan at the very start or the 3D picture later
Reply 39
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman

... is the z^2=1-x a plane or just a curve?



a curve

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you