Question about finding volume using integrationI need to solve the volume that's between: z = 0 , x = 0 , y = 0 , x 2 + y 2 = 4 z = 12 − x − y1) Does it matter if I use double integrals or triple?2) When I draw this area on the x and y plane, I have a circle with radius 2. How do I know which quadrant my volume is in?

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Question about finding volume using integrationI need to solve the volume that&#039;s between:  z  =  0  ,    x  =  0  ,    y  =  0  ,        x    2    +      y    2    =  4    z  =  12  −  x  −  y1) Does it matter if I use double integrals or triple?2) When I draw this area on the x and y plane, I have a circle with radius 2. How do I know which quadrant my volume is in?

Momellaxi · Accepted Answer

Step 11) No. You can set this up as either:      ∫    0    2        ∫    0                  4        −                  x          2                      z  d  x  d  y  =      ∫    0    2        ∫    0                  4        −                  x          2                      (  12  −  x  −  y  )  d  y  d  xWhich represents integrating the volume bounded by the cylinder       x    2    +      y    2    =  4 bounded above by the plane  12  −  x  −  y  =  zBut this is just a shortcut for       ∫    0    2        ∫    0                  4        −                  x          2                          ∫    0          12      −      x      −      y        d  z  d  y  d  x.Step 22) You technically can&#039;t tell which octant you are in, but I think what was intended is as follows.Assuming you are in the first octant you have   0  ≤  z  ≤  12  −  x  −  y    0  ≤  y  ≤      4    −          x      2          0  ≤  y  ≤  2

I need to solve the volume that's between: z=0, x=0, y=0, x^{2}+y^{2}=4. z=12-x-y.

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