Volume of Solid Enclosed by an EquationI'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:Compute the volume of the solid enclosed by1. x a + y b + z c = 1 , x = 0 , y = o , z = 02. x 2 + y 2 − 2 a x = 0 , z = 0 , x 2 + y 2 = z 2

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Volume of Solid Enclosed by an EquationI&#039;m having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:Compute the volume of the solid enclosed by1.       x    a    +      y    b    +      z    c    =  1  ,  x  =  0  ,  y  =  o  ,  z  =  02.       x    2    +      y    2    −  2  a  x  =  0  ,  z  =  0  ,      x    2    +      y    2    =      z    2

Macie Melton · Accepted Answer

Step 1The first equation is equivalent to  (  x  −  a      )    2    +      y    2    =      a    2    ,, which is a cylinder centered at (a,0,0) parallel to the z-axis. It&#039;s intersection with the xy-plane is a circle; the volume we want is the solid above this circle and below       z    2    =      x    2    +      y    2  . Let the region bounded by this circle be C, so the volume is       ∬    C              x      2        +          y      2          d  A  =  2      ∫    0                  π        2                  ∫    0          2      a      cos      ⁡      θ            r    2      d  r    d  θ  ,, where we used polar coordinates, in which       r    2    =      x    2    +      y    2   and   d  A  =  r    d  r    d  θ (loosely speaking). The bounds come from the fact that half of the region C is also described by the polar curve   r  =  2  a  cos  ⁡  θ, where   0  ≤  θ  ≤      π    2  , and the multiplication by 2 accounts for the other half. Then, evaluating this, we get  2      ∫    0                  π        2                        [                        r          3                3            ]        0          2      a      cos      ⁡      θ          d  θ  =            16              a        3              3        ∫    0                  π        2                  cos    3    ⁡  θ    d  θ  .Step 2This last integral can be evaluated using standard techniques, and turns out to be       2    3   for a final answer of             32              a        3              9

Compute the volume of the solid enclosed by: 1. x/a+y/b+z/c=1, x=0, y=0, z=0. 2. x^2+y^2-2ax=0, z=0, x^2+y^2=z^2.

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