I am given the following integral:
$$\int_0^2\int_0^{y^3}\int_0^{y^2}f(x,y,z) dzdxdy $$
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2f(x,y,z)dydzdx\,+\,\int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split?
Best Answer
The region of the integral is characterized by $$ 0 \leq y \leq 2 \\ 0 \leq x \leq y^3 \\ 0 \leq z \leq y^2 $$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 \leq x \leq y^3 \leq 2^3 = 8 \\ 0 \leq z \leq y^2 \leq 4$$
The $0 \leq x \leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely $$ y \geq x^{1/3} \\ y \geq z^{1/2}$$ And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $z\leq x^{2/3}$ and the limits will be $$ \int_{x=0}^8\int_{z= 0}^{x^{2/3}} \int_{y=x^{1/3}}^2 $$ and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z \leq y^2 \leq 4$ and the limits will be $$ \int_{x=0}^8\int_{z= x^{2/3}}^{4} \int_{y=z^{1/2}}^2 $$