# Setting up the triple integrals for a solid given by $y+z=2$ and $x=4-y^2$?

I'm trying to set up all six triple integrals to find the volume of the solid that lies in the first octant bounded by the coordinate planes, the plane $$y+z=2$$, and the cylinder $$x=4-y^2$$.

$$3$$D-Graph:

I've been able to set up the triple integrals for every other combination except for $$dydxdz$$ and $$dydzdx$$, which I am struggling to find the bounds for.

I know the volume of the solid is $$20/3$$, but no matter what I try I haven't been able to come up with $$dydxdz$$ and $$dydzdx$$ to produce that result. Looking at the $$3$$D graph, the projection onto the $$xz$$-plane appears to be the equation $$z=2-(4-x)^{1/2}$$ and $$x=4$$ (see image below), but I don't think that is correct since it isn't producing the correct volume.

$$xz$$-plane projection?

If that isn't correct, what does the projection look like?

I'd appreciate any help. Thank you!

• It would be nice if those voting to close this question as off-topic would shed some light on what other context or details should OP provide that they haven't already. – Ennar Nov 5 '18 at 0:15

## 1 Answer

The boundary of the solid is given by the following inequalities:

$$x,y,z\geq 0,\\ y + z \leq 2,\\ x+y^2\leq 4.$$

If you let $$y = 0$$ in the above, it follows $$0\leq x\leq 4$$ and $$0\leq z\leq 2$$, so the projection onto $$xz$$-plane is rectangle $$[0,4]\times[0,2]$$.

So, what remains is to express the boundary of the solid directly above the rectangle as function $$y=f(x,z)$$. We have two conditions on $$y$$:

$$0\leq y\leq 2-z,\\ 0\leq y\leq \sqrt{4-x},$$

and both of them need to be satisfied. Thus, $$0\leq y \leq \min\{\sqrt{4-x},2-z\}$$ and therefore, you need to integrate

$$\int_0^4\int_0^2\int_0^{\min\{\sqrt{4-x},2-z\}}1\,dydzdx.$$

Now, you might be thinking what in the world to do with that. But before I explain how to do it, let me digress with analogous $$2$$-dimensional problem.

Let's say that we want to calculate area of the shape in the first quadrant, bounded by $$y=x$$ and $$y = 1 -x^2$$.

One way to do it is to write the area as the integral $$\int_0^{\frac{-1+\sqrt 5}2}\int_y^{\sqrt{1-y}}1\,dxdy$$ but you can also do it this way

$$\int_0^1\int_0^{\min\{x,1-x^2\}}1\,dydx$$

as I'm sure you've seen before. What? No? Ok, ok, I'll stop joking around.

We have inequalities $$x,y\geq 0,\\ y\leq 1-x^2,\\ y\leq x.$$ If you let $$y = 0$$ in the above, you get $$0\leq x\leq 1$$, so the segment $$[0,1]$$ is the projection onto $$x$$-coordinate. We should express $$y$$ as function of $$x$$, and from the inequalities, we get $$0\leq y \leq \min\{x,\sqrt{1-x^2}\}$$, as I claimed above.

But this is not how we solve this usually. What you do is split the integral into two pieces:

$$\int_0^{\frac{-1+\sqrt 5}2}\int_0^x1\,dydx + \int_{\frac{-1+\sqrt 5}2}^1\int_0^{1-x^2}1\,dydx.$$

This is the same thing as the last integral. We split the segment $$[0,1]$$ into $$[0,\frac{-1+\sqrt 5}2]$$ and $$[\frac{-1+\sqrt 5}2,1]$$. This is because

$$\min\{x,\sqrt{1-x^2}\}=\begin{cases} x, & x\in [0,\frac{-1+\sqrt 5}2],\\ 1-x^2, & x\in [\frac{-1+\sqrt 5}2,1]. \end{cases}$$

Back to the original problem.

So, learning from $$2$$-dimensional case, we now know that we need to separate the rectangle $$[0,4]\times [0,2]$$ in the $$xz$$-plane into two parts: one where $$\sqrt{4-x}\leq 2-z$$ and the other where $$2-z\leq \sqrt{4-x}$$. We can equate those two things to get a curve given by $$\sqrt{4-x} = 2 - z$$. It looks like this:

In the green area we have $$\sqrt{4-x}\leq 2-z$$ and in the blue area we have $$2-z\leq \sqrt{4-x}$$. Therefore,

\begin{align} \int_0^4\int_0^2\int_0^{\min\{\sqrt{4-x},2-z\}}1\,dydzdx &= \color{green}{\int_0^4\int_0^{2-\sqrt{4-x}}}\int_0^{\sqrt{4-x}}1\,dydzdx + \color{blue}{\int_0^4\int_{2-\sqrt{4-x}}^2}\int_0^{2-z}1\,dydzdx\\ &= \color{green}{\int_0^2\int_{4-(z-2)^2}^4}\int_0^{\sqrt{4-x}}1\,dydxdz+\color{blue}{\int_0^2\int_0^{4-(z-2)^2}}\int_0^{2-z}1\,dydxdz\\ &=\frac{20}3. \end{align}