Changing order of $\int_{-23/4}^4\int_0^{4-y}\int_0^{\sqrt{4y+23}} f(x,y,z) \,dx\,dz\,dy$ to $\iiint f(x,y,z) \, dy\,dz\,dx$

Question

The sketch of the region is:

Based on it I get:

$$\int_0^{\sqrt{23}}\int_0^{4-y}\int_{\frac{x^2-23}{4}}^{4} f(x,y,z)\, dy\,dz\,dx$$

But this arrangement makes me drag $y$ so I must be missing something.

Note: I am just interested in $\iiint f(x,y,z) \, dy\,dz\,dx.$

Answer 1

The first part I notice with your solution is you have an integral with an upper limit of y after you already integrated with respect to y. Once you have integrated with respect to a given variable, limits beyond that should not include that variable for multiple reasons. Firstly, if another variable really does depend on y, and you integrated with respect to y and treated that other variable as constant, then clearly this is incorrect. Secondly, from a pure practical perspective you would end up with a function of y instead of a numerical answer in the end.

For the question at hand, y will have its upper bound be a function of z and lower bound a function of x. The other two variables, z and x, vary independently so their limits will be constants. Can you finish the question from that?