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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!

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  • $\begingroup$ 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. $\endgroup$ – Ennar Nov 5 '18 at 0:15
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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 $y = 1 -x^2$.

enter image description here

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:

enter image description here

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}

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