# Bounds on triple integral (Cartesian)

I want to setup a triple integral for the volume of the surface in the ordering $$dy \hspace{1mm} dx \hspace{1mm} dz$$:

So far I have that for $$0\leq z \leq 1, 0 \leq y \leq x$$ and for $$1 \leq z \leq 2, 0 \leq y \leq \sqrt{2-z}$$. I'm having trouble setting up bounds for $$x$$. It looks from the projection like $$0 \leq x \leq 1$$ for both integrals, but it doesn't give me the right value for volume (should be $$\frac{11}{12}$$ based on the other differential orderings.)

• You'll need three triple integrals with the ordering $dxdydz$
– user801306
Commented Mar 7, 2021 at 15:57

For $$0 \leq z \leq 1$$, the volume is simply bound by $$y = 0$$, $$y = x$$ and $$x=1$$.

For any given value of $$z \in (1,2)$$, there are two bounds of $$y$$: below $$x = \sqrt{2-z}$$, it is simply bound by plane $$y = x$$ and above $$x = \sqrt{2-z}$$, by the parabolic cylinder.

So the integral should be

$$\displaystyle \int_0^1 \int_0^1 \int_0^x \ dy \ dx \ dz$$ +

$$\displaystyle \int_1^2 \int_0^{\sqrt{2-z}} \int_0^x \ dy \ dx \ dz$$ +

$$\displaystyle \int_1^2 \int_{\sqrt{2-z}}^1 \int_0^{\sqrt{2-z}} \ dy \ dx \ dz$$

• Let me know if you are ale to visualize it looking at the bounds now and if it is clear to you. Commented Mar 7, 2021 at 16:24
• Thank you! Is $x=\sqrt{2-z}$ the projection on the xz plane? Since this is true only when y = x, I pictured this curve off of the xz plane (at some positive y value), somehow touching the y = x plane. Commented Mar 9, 2021 at 2:55
• In the picture you have attached with your question, you notice there is a dotted line at $z=1$. That is the upper bound of the triangular prism - first integral. Above $z = 1$, You can see a curve in the plane $y=x$. That is intersection of $z \leq 2-y^2$ and the plane. Given the curve is in plane $y=x$, we have $z=2-x^2 = 2-y^2$. At any $z$, all values of $x$ where $x \leq \sqrt{2-z}$, you can see in the picture that, a shell along $y$ will be bound by the plane $y=x$ whereas beyond that curve ($x \geq \sqrt{2-z}$), you can see parabolic cylinder surface and that is the upper bound of $y$. Commented Mar 9, 2021 at 5:26