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Thread: surface integral

  1. #1
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    surface integral

    By using the double integral , find surface area of the portion of the surface z = sqrt (4 -x^2) that lies above rectangle in xy -plane whose coordinate dsatisfy 0<x<1 and 0<y<4 ...

    I have sketched out the diagram , but i found that , it's merely 2 plane surfaces , it's not a 3d object formed by 2 surfaces , right ? So, we just need to find the respective surface area and add 2 things up ?

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  2. #2
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    Re: surface integral

    Hey xl5899.

    For this object you will need to let y vary and when you sketch it you should get a cylinder like object where the cross section circle is in the x-z plane instead of the x-y plane [where it usually is for a normal cylinder].

    Try sketching that to see what it looks like [set y = 0 for the first circle and then draw the same circle across the y-axis to get the cylinder like surface you need to evaluate].
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    Re: surface integral

    Quote Originally Posted by chiro View Post
    Hey xl5899.

    For this object you will need to let y vary and when you sketch it you should get a cylinder like object where the cross section circle is in the x-z plane instead of the x-y plane [where it usually is for a normal cylinder].

    Try sketching that to see what it looks like [set y = 0 for the first circle and then draw the same circle across the y-axis to get the cylinder like surface you need to evaluate].
    do you mean copy the same set of xz plane i have now to y = 4 and join the lines between them to form 3d object ?
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