# How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and the plane y = 2 - 2z? I need to sketch and label...

How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and the plane y = 2 - 2z?

I need to sketch and label the volume V as well.

`int` `int` `int` dx dy dz

`int` `int` `int` dy dx dz

`int` `int` `int` dz dy dx

Borys Shumyatskiy | Certified Educator

First octant means `xgt=0, ygt=0, zgt=0.` The formula `y=27/8 x^3` may be written as `y=(3/2 x)^3.` We need the (simple) inverse formulas, too: `x=2/3 root(3)(y), z=1-y/2.`

Because `zgt=0` we see `ylt=2` and therefore `xlt=2/3 root(3)(2).` Also, `z=1-y/2lt=1.` Now we are ready to write the integral in different orders of variables.

If `x` is fixed, `y` can be between `(3/2 x)^3` and `2` in `xy` plane `z=0.` For fixed `x` and `y` we see `z` can be between `0` and `1-y/2,` i.e. the integral becomes

`int_(x=0)^(x=2/3 root(3)(2)) dx int_(y=(3/2 x)^3)^(y=2) dy int_(z=0)^(z=1-y/2) dz(1) = int_(x=0)^(x=2/3 root(3)(2)) int_(y=(3/2 x)^3)^(y=2) int_(z=0)^(z=1-y/2) 1 dz.`

If we start from `y` it can be from `0` to `2,` then `x` can be from `0` to `2/3 root(3)(y)` and `z` can be from `0` to `1-y/2.` The integral is

`int_(y=0)^(y=2) int_(x=0)^(x=2/3 root(3)(y)) int_(z=0)^(z=1-y/2) 1 dy dx dz.`

And if we start with `z,` it can be from `0` to `1,` then `y` can be from `0` to `2-2z` and `x` from `0` to `2/3 root(3)(y),` i.e.

`int_(z=0)^(z=1) int_(y=0)^(2-2z) int_(x=0)^(x=2/3 root(3)(y)) 1 dz dy dx.`

It is not that simple to sketch `V` but I tried. :)

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