Reversing order of triple integral

Question

I have a question where I need to reverse the order of integration. But I'm struggling to understand how exactly to do so.

Any help would be appreciated. How does the equation x=yz look like?

Answer 1

First we need to find the bounds on $x$. The lower bound is given as $0$ and the upper bound happens when $x=yz$. For a given value of $z$, the biggest value of $y$ is $3-z$, so the $x\le(3-z)z$. Differentiating, $$\frac d{dz}(3-z)z=3-2z=0$$ So $z=\frac32$ and $x\le(3-\frac32)\frac32=\frac94$ gives us the upper bound for $x$. For a given $x$, we know that $x\le yz$, so $z\ge \frac xy$. Also $y\le3-z$, so $z\le3-y$. Thus $z$ lies above the hyperbola $z=\frac xy$ and below the line $z=3-y$. They cross when $3-y=\frac xy$ or $y^2-3y+x=0$, that is when $$y=\frac{3\pm\sqrt{9-4x}}2$$ So this tells us the limits on $y$ and we already have the limits on $z$, so $$\int_0^3\int_0^{3-z}\int_0^{yz}f(x,y,z)dx\,dy\,dz=\int_0^{\frac94}\int_{\frac{3-\sqrt{9-4x}}2}^{\frac{3+\sqrt{9-4x}}2}\int_{\frac xy}^{3-y}f(x,y,z)dz\,dy\,dx$$