Rewrite the integral $$\int_0^3 \int_0^{9-y^2} \int_\frac{y}3^1 f(x,y,z) dx dz dy $$ as an interated integral in the order dy dz dx.
I have trouble visualizing if my answer is the correct iterated integral or not. I got$$\int_0^1 \int_0^{9-9x^2} \int_0^{3x} f(x,y,z) dy dz dx $$ but i am unsure if this is correct or perhaps there is some part of the domain not captured in my answer.
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$\begingroup$If you project the integration region in the $xz$-plane you'll see that the integral must be split in two:
$$ \int_0^1 \int_0^{9-9x^2}\int_0^{3x}f(x,y,z) dy dz dx + \int_0^1 \int_{9-9x^2}^9\int_0^{\sqrt{9-z}}f(x,y,z) dy dz dx. $$
$\endgroup$ $\begingroup$At the intersection of plane $y = 3x$ and parabolic cylinder $z = 9 - y^2$, we have$$z = 9 - 9x^2$$
So for $~0 \leq z \leq 9 - 9x^2, ~y$ is bound above by the plane $y = 3x$. This is the region that your integral covers. But it misses the region above $z = 9 - 9x^2$.
For $z \geq 9 - 9x^2$, $y$ is bound above by the parabolic cylinder and not by the plane $y = 3x$. So, the limits should be $0 \leq y \leq \sqrt{9-z}$. The integral is as shown below,
$ \displaystyle \int_0^1 \int_{9-9x^2}^9 \int_0^{\sqrt{9-z}} f(x,y,z)~ dy ~dz ~dx$
You can replace $f(x, y, z)$ with $1$ to evaluate both integrals that should give you the volume of the region. You can confirm if the answers are same in both cases or not. Using WolframAlpha, I find the answer as $ \displaystyle \frac{45}{4}$.
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