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Math Help - Cylindrical Integration

  1. #1
    Newbie sherryl's Avatar
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    Cylindrical Integration

    I want to evaluate ∫∫∫√(x+y) dV over the region in the first octant bounded by

    x+y=2
    x+y=√(2) x
    x+y=z (-->paraboloid of revolution)

    Here's how I've set up my cylindrical coordinates:

    \int\nolimits_{0}^{\tfrac{\pi }{2}} {d\theta }\int\nolimits_{\sqrt {2} \cos \theta }^{\sqrt {2} } {r^{2}dr}\int\nolimits_{0}^{r^{2}} {dz}

    Could somebody please confirm that this is the correct approach?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by sherryl View Post
    I want to evaluate ∫∫∫√(x+y) dV over the region in the first octant bounded by

    x+y=2
    x+y=√(2) x
    x+y=z (-->paraboloid of revolution)

    Here's how I've set up my cylindrical coordinates:

    \int\nolimits_{0}^{\tfrac{\pi }{2}} {d\theta }\int\nolimits_{\sqrt {2} \cos \theta }^{\sqrt {2} } {r^{2}dr}\int\nolimits_{0}^{r^{2}} {dz}

    Could somebody please confirm that this is the correct approach?
    This is correct. Clearly you found the correct answer by yourself, but just incase this helps I'm going to show you a tip (if you didnt know it already).

    When we are given

     2= x^2 + y^2 and  \sqrt{2} x = x^2 + y^2 we are given a region in the xy domain bounded by a circle of radius  \sqrt{2} and by a circle of some other unknown radius (we need to figure this out). Further, we should note that the second circle is actually shifted on the x-axis, so to properly conclude what radius we need in our integral, we should draw this all out.

    What we can do is complete the square,

     \sqrt{2} x = x^2 + y^2

     0 = x^2 - \sqrt{2}x + y^2

     0 = (x - \frac{ \sqrt{2} }{2})^2 - \frac{1}{2} + y^2

     \frac{1}{2} = (x - \frac{ \sqrt{2} }{2})^2 + y^2

    Now we can see our radius clearly. Our radius is  \frac{1}{ \sqrt{2} } and the circle is centered at  ( \frac{ \sqrt{2} }{2} , 0 )

    If we add our radius plus our shift along the x-axis, we can find the distance this circle travels as to see if it lies inside or outside the radius of the circle centered at the origin.

    It so happens that  \frac{ \sqrt{2} }{2} + \frac{1}{ \sqrt{2} } = \sqrt{2} so we cant tell much from this, other then that our circles in the xy domain overlap along the x-axis at the very most right end. But if you select a value of x that is minimally smaller then  \sqrt{2} and sub it into both equations, you will find that the radius of the second circle is now less then  \sqrt{2} . Thus, we are going from the inside of our shifted circle to the edge of our regular circle

    To most apply model this we can go back to the original equation and switch to polars,

     \sqrt{2} x = x^2 + y^2

     \sqrt{2} rcos \theta = r^2 \to r = \sqrt{2} cos \theta

    Of course leading to,

     \int_{ 2 cos \theta }^{ \sqrt{2} }
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