determine volume of region from description of views

Smily

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May 27, 2006
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can someone help me, please? :roll:

A region in space, when viewed from three different views, looks like a circle, a square, and a triangle. Describe this object. Use multiple integration to determine its volume, in terms of the diameter a

Thank you!
 
Re: determine its volume

Hello, Smily!

A region in space, when viewed from 3 different views,
looks like a circle, a square, and a triangle.
Describe this object.
Use multiple integration to determine its volume, in terms of the diameter a\displaystyle a

We have a right circular cylinder.
The base is a circle of diameter a\displaystyle a; its height is a.\displaystyle a.
. . Viewed from above, it looks like a circle.
. . Viewed from the side, it looks like a square.

Looking from the "front", slice the cylinder so it looks like an isosceles triangle.
Code:
      ......*......
      :    /:\    :
      :   / : \   :
      :  /  :a \  :
      : /   :   \ :
      :/    :    \:
      *-----------*
              a/2

Place the object on the xy-plane, centered at the origin.
Due to its symmetry, we can find the volume in the first octant and multiply by 4.

The volume is above z=0\displaystyle z\,=\,0 and below the plane z=a2y\displaystyle z \:=\:a\,-\,2y

The base is the circle: x2+y2=(a2)2\displaystyle \,x^2\,+\,y^2\:=\:\left(\frac{a}{2}\right)^2
So y\displaystyle y ranges from y=0\displaystyle y\,=\,0 to y=12a24x2\displaystyle y\:=\:\frac{1}{2}\sqrt{a^2\,-\,4x^2}

Finally, x\displaystyle x ranges from x=0\displaystyle x\,=\,0 to x=a2\displaystyle x\,=\,\frac{a}{2}


. . . . . . . . . . . . . . . . . . . . . . . . . a2    12a24x2  a2y\displaystyle _{\frac{a}{2}}\;\;_{\frac{1}{2}\sqrt{a^2-4x^2}}\;\,_{a-2y}
I think the volume is: \(\displaystyle \L\:V \;= \;4\int_0\int_0\;\;\;\;\;\int_0\;\;dz\,dy\,dz\)

. . but don't quote me . . .

 
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