Question

Rewrite the following in dx dz dy order.

Answer 1

`x` is minimized at
`x=y^2`, so the upper bound for
`z` can be rewritten as

`z = 2 - \frac x2 = 2 - \frac{y^2}2`

so that in the integral with
`dz`, we have the range
`0 \le z \le 2 - \frac{y^2}2`.

Rewrite the plane equation as a function
`x=x(z)`.

`z = 2 - \frac x2 \implies 2z = 4 - x \implies x = 4 - 2z`

So the equivalent integral is the one in choice A,

`\displaystyle \int_(-2)^2 \int_(y^2)^4 \int_0^(2-x/2) dz\,dx\,dy = \boxed{\int_(-2)^2 \int_0^(2-y^2/2) \int_(y^2)^(4-2z) dx\,dz\,dy}`