Let V be volume bounded by surface y 2 = 4 a x and the planes x + z = a and z = 0.Express V as a multiple integral, whose limits should be clearly stated. Hence calculate V.SolutionI want to find out the limits of the multiple integral that's needed to calculate V.I'm guessing that: x = a − z, x = ( y 2 ) / 4 a. y = ± 4 a x , z = 0 , z = a − xBut it seems like I've used the upper plane equation twice?Also, it would really help if we could compare our answer volumes to check that this is right from the start!

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Let V be volume bounded by surface       y    2    =  4  a  x and the planes   x  +  z  =  a and   z  =  0.Express V as a multiple integral, whose limits should be clearly stated. Hence calculate V.SolutionI want to find out the limits of the multiple integral that&#039;s needed to calculate V.I&#039;m guessing that:   x  =  a  −  z,   x  =  (      y    2    )      /    4  a.  y  =  ±      4    a    x    ,    z  =  0  ,    z  =  a  −  xBut it seems like I&#039;ve used the upper plane equation twice?Also, it would really help if we could compare our answer volumes to check that this is right from the start!

ambivalentnoe1 · Accepted Answer

Step 1OK so first of all we must turn &quot;bounded by&quot; into inequalities. If you picture the three surfaces, you realize the only bounded region is:      {          y      2        ≤    4    a    x    ,    z    ≥    0    ,    x    +    z    ≤    a    }    =      {    y    ∈    [    −          4      a      x        ,          4      a      x        ]    ,    x    ,    z    ≥    0    ,    x    ≤    a    −    z    }    .The limits for y are explicit and depend on x, so we put the dy integral inside. With that, x,z don&#039;t depend on y, so we will have one from 0 to a, and the other in such a way that the sum is less than a. I wrote the equation in terms of x to suggest my approach, but swapping the integrals shouldn&#039;t give any change.Step 2How do we see that is the region? Well we have two planes and a &#039;parabolc prism&#039;. The region outside the prism will be unbounded, which for example gives   x  ≥  0 and the y inequality. Under the   z  =  0 plane, the region can go to infinity in x, since there is no upper bound for it from any of the surfaces. So   z  ≥  0. The other bit,   x  ≤  a  −  z, is just the bounding of the plane, which has to be an upper bound since the other plane is a lower one and the &#039;prism&#039; doesn&#039;t touch z in any way.

Let V be volume bounded by surface y^2=4ax and the planes x+z=a and z=0. Express V as a multiple integral, whose limits should be clearly stated. Hence calculate V.

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