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Math Help - Intersection of paraboloid and plane

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    Intersection of paraboloid and plane

    If C is the curve in reals^3 which is the intersection of the paraboloid {(x,y,z) s.t. z= (x^2) + (y^2)}
    With the plane {(x,y,z) s.t. z= 3 -2y}
    And L is the segment of C which lies in the half-space x>or=0. What is integral fds with bottom limit L (on integral) where f(x,y,z) := 5xy +xz?
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    Quote Originally Posted by maximus101 View Post
    If C is the curve in reals^3 which is the intersection of the paraboloid {(x,y,z) s.t. z= (x^2) + (y^2)}
    With the plane {(x,y,z) s.t. z= 3 -2y}
    And L is the segment of C which lies in the half-space x>or=0. What is integral fds with bottom limit L (on integral) where f(x,y,z) := 5xy +xz?
    If I am reading this correctly then the first thing you need to find is C.
    IF you subract the equation of the plane from the equation of the paraboloid we get

    \begin{array}{rrrr} z=&x^2&+y^2& \\ -z=&& +2y&-3 \\ 0=& x^2&+y^2+2y&-3\end{array}
    Or
    x^2+y^2+2y=3 \iff x^2+(y+1)^2=4

    This is a circle of radius 2 in the xy plane and can be parametrized as x=2\cos(t) and y=-1+2\sin(t)

    Now if we put this into the equation of the plane we get

    z=3-2y=3-2(-1+2\sin(t))=5-4\sin(t)

    So now we have a parametrization of the needed curve
    \vec{r}(t)=[2\cos(t)]\vec{i}+[-1+2\sin(t)]\vec{j}+[5-4\sin(t)]\vec{k}<br />

    Can you finish from here?
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    thank you that helped a lot, so I found f(x,y,z) = 5xy+xz= 12cos(t)sin(t) and ds=2sqrt(1+4(cost)^2) dt

    so what are the limits for my integral? in finding ∫ [L] f(x,y,z)ds and could you explain how we found them.
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    Quote Originally Posted by maximus101 View Post
    And L is the segment of C which lies in the half-space x>or=0. [/SIZE][/FONT]
    so what are the limits for my integral?
    Well x=2\cos(t) so 2\cos(t) \ge 0 \implies ...
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    so bounds are -pi/2 lower and pi/2 upper
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    Yes
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    cool thank you
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