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    (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

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    (Original post by cooldudeman)
    Calculate the work done by the force field F(x,y,z)=(y^2,z^2,x^2) along the curve of intersection of the sphere x^2+y^2+z^2=1, the cylinder x^2+y^2=x, and the halfspace z>0. The path is traversed in a direction that appears clockwise when viewed from the high above the xy-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:

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

    where x \in [0,1]

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

    \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)

    \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

    \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 \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.
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    (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:

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

    where x \in [0,1]

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

    \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)

    \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

    \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 \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?
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    (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:

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

    where x \in [0,1]

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

    \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)

    \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

    \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 \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
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    (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:

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

    where x \in [0,1]

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

    \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)

    \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

    \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 \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.

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    (Original post by TeeEm)
    ignore me

    I just saw the initial comment
    I'm just confused on this.

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    (Original post by cooldudeman)
    I'm just confused on this.

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    not quite that

    z>0 and 0<x<1
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    (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??

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    (Original post by cooldudeman)
    Ok if thats the case then why isn't it just the top right quadrant??

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    it is octant 1 and octant 4
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    (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?

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    (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?

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    octant is 1/8 of a sphere

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

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    (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)?

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    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.
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    (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.

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    (Original post by cooldudeman)
    I see so how did you come up with the limits? Hpw did you just know them.

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    I will show you in an isolated example in about an hour.
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    (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.

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    (Original post by cooldudeman)
    OK thanks. I think I understand everything beaides how you found the limits of t.

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    see if this helps

    IMG.pdf
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    (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?
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    (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
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    (Original post by cooldudeman)
    ... is the z^2=1-x a plane or just a curve?

    a curve
 
 
 
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