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

$z = 2 - \dfrac x2 = 2 - \dfrac{y^2}2$

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 $x=x(z)$.

$z = 2 - \dfrac x2 \implies 2z = 4 - x \implies x = 4 - 2z$

So the equivalent integral is the one in choice A,

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