I need to calculate some volumes, but I can't because I'm not able to visualize them.I 'm not able to set the integrals. I'm going to show my problem:
I have that volume $V:=\{\ \underline{x} \in \mathbb{R^3} : x,y,z \ge 0 ,\ z \le 2-x^2-y,\ z \le x+2y \} \\ \underline{x}:=(x,y,z)$
How can I see it? Should I only use algebra and abandon any visual approach? If yes how? Thanks in advance
P.S. Obviously the tools like Geogebra 3D I don't take them into consideration, I would like to succeed in this without using these tools.
I would also like to add that generally if I can see the volume I have no problems with the integral calculation. The problem is all there, sometimes I can see the volume at other times absolutely not!
I assume throughout much of this that you're comfortable with at least double integrals and related iterated integrals.
First I discuss how to visualize the region enough to set up the integrals, focusing primarily on understanding the "shadow" in the $xy$-plane. Then, after a second horizontal line, I also show a wordy algebra-only calculation to set up the integrals.
How to visualize a volume:
Boundary Surfaces (skippable):
We might like to visualize the boundary surfaces $x=0$, $y=0$, $z=0$, $z=2-x^{2}-y$ and $z=x+2y$. It will turn out that visualizing these 3D graphs is not really needed for a problem like this.
The first three, $x=0$, $y=0$, $z=0$, are just the coordinate planes.
$z=x+2y$ is a plane (because it's a degree $1$ equation) that passes through the origin with cross-sections that have slope $1$ in the $x$ direction (for fixed $y$) and $2$ in the $y$ direction (for fixed $x$). (It's also the case that $\left\langle 1,2,-1\right\rangle$ is perpendicular to the plane, but I don't find that as helpful for this sort of visualization.)
It remains to understand $z=2-x^{2}-y$. There are a lot of options, but I still like cross-sections here. For each fixed $y$, we have a parabola. For example, $y=0$ gives the downward parabola $z=2-x^{2}$. $y=1$ gives the downward parabola $1-x^{2}$. And $y=2$ gives the downward parabola $z=-x^{2}$, which never has $z$ positive. This already tells us that all of the points of the volume satisfy $0\le y\le2$, as $y>2$ would mean $0\le z\le2-x^{2}-y\le2-y<0$. For each fixed $x$, we have a line. For example, $x=0$ gives $z=2-y$. $x=1$ gives $z=1-y$. And $x=\sqrt{2}$ gives $z=-y$, which never has $z$ positive for $y\ge0$. So all points of the volume satisfy $0\le x\le\sqrt{2}$.
Intersections:
Whether or not we understand the boundary surfaces, we need to understand where they intersect within the volume.
Intersections of Surfaces:
As long as we keep in mind that we're in the first quadrant where $x,y\ge0$, we don't have to consciously worry about the intersections with $x=0$ and $y=0$ right now, since the rest of our conditions are given in terms of $z$, which means we're likely to set up a double integral with respect to $x$ and $y$.
Let's start with $z=0$. $z=x+2y$ intersects $z=0$ when $x+2y=0$ so that $y=-x/2$. But since we have $x,y\ge0$, we won't run into $y=-x/2$ except at the origin. And $z=2-x^{2}-y$ intersects $z=0$ when $y=2-x^{2}$, which makes a parabola in the $xy$-plane (as opposed to the vertical, parallel to the $xz$-plane, parabolas discussed above).
Now, it matters which side of the parabola $y=2-x^{2}$ has a piece of the volume. If $y>2-x^{2}$, then $z=2-x^{2}-y<0$, so that region is not part of the volume. It's only the region of the first quadrant of the $xy$-plane where $y\le2-x^{2}$ that matters.
Finally, above where in the $xy$-plane do the two complicated surfaces intersect? $z=2-x^{2}-y=x+2y$ means $y=\dfrac{2-x-x^{2}}{3}$ (a different parabola). Again, it matters what happens on both sides of this parabola. If $y<\dfrac{2-x-x^{2}}{3}$, then $x+2y$ decreases and $2-x^{2}-y$ increases, so that $z\le x+2y$ is the stronger condition. On the other hand, if $y>\dfrac{2-x-x^{2}}{3}$, then $z\le2-x^{2}-y$ is the stronger condition.
Intersections of curves:
You might think we're done, especially if you've graphed things with a computer/calculator already. But we need to seriously consider what happens with these two parabolas in the $xy$-plane. It matters which one is outside the other, or if they intersect in the first quadrant. $y=\dfrac{2-x-x^{2}}{3}$ and $y=2-x^{2}$ intersect at the points $\left(\dfrac{1\pm\sqrt{33}}{4},\dfrac{-1\mp\sqrt{33}}{8}\right)$. One of those two points has a negative value for $x$, and the other has a negative value for $y$, so neither appears in the first quadrant. Thus, one parabola is outside the other in the first quadrant.
If $y=\dfrac{2-x-x^{2}}{3}$ were the outside one, then the intersection of $z=x+2y$ and $z=2-x^{2}-y$ wouldn't matter; it would happen beyond the curve where $z=2-x^{2}-y$ dips below $z=0$. But actually, by considering some test ray like $x=y$, we see that $y=\dfrac{2-x-x^{2}}{3}$ includes $\left(\sqrt{6}-2,\sqrt{6}-2\right)$ and $y=2-x^{2}$ includes $\left(1,1\right)$. So $y=2-x^{2}$ is the outer parabola.
Picture and integrals:
Picture:
Putting all of the intersection information together, we arrive at a schematic of the first quadrant that looks like this:
The region labeled "N/A" does not lie under the volume. The other two regions are for $2-x^{2}-y$ and $x+2y$ representing the height above the points in those regions.
Integrals:
We know the $y=2-x^{2}$ parabola intersects the axes at $(0,2)$ and $\left(\sqrt{2},0\right)$. In order to actually set up a sum of iterated integrals, we need to know (at least one of) the intersections of $y=\dfrac{2-x-x^{2}}{3}$ with the axes as well. They are $\left(0,\frac{2}{3}\right)$ and $(1,0)$.
This leads naturally to a sum of three iterated integrals with $x$ on the outside:
$$\int_{0}^{1}\left(\int_{0}^{\dfrac{2-x-x^{2}}{3}}x+2y\,\mathrm{d}y+\int_{\dfrac{2-x-x^{2}}{3}}^{{\displaystyle 2-x^{2}}}2-x^{2}-y\,\mathrm{d}y\right)\,\mathrm{d}x$$ $$+\int_{1}^{\sqrt{2}}\int_{0}^{{\displaystyle 2-x^{2}}}2-x^{2}-y\,\mathrm{d}y\,\mathrm{d}x$$
However, using the $\left(0,\frac{2}{3}\right)$ point and solving the parabola equations for $y$, we could also write:
$$\int_{0}^{2/3}\left(\int_{0}^{\left(-1+\sqrt{9-12y}\right)/2}x+2y\,\mathrm{d}x+\int_{\left(-1+\sqrt{9-12y}\right)/2}^{\sqrt{2-y}}2-x^{2}-y\,\mathrm{d}x\right)\,\mathrm{d}y$$ $$+\int_{2/3}^{2}\int_{0}^{\sqrt{2-y}}2-x^{2}-y\,\mathrm{d}x\,\mathrm{d}y$$
Checking our answers:
We can use a computer to check our answers. For instance:
Wolfram Language (Mathematica), 415 bytes
Try it online!
The output is
{-17/60 + (16*Sqrt[2])/15, -17/60 + (16*Sqrt[2])/15, -17/60 + (16*Sqrt[2])/15}, indicating that whether you use Mathematica's black box calculation of the volume, or either iterated integral setup, you always get $\dfrac{16\sqrt2}{15}-\dfrac{17}{60}$.Only using algebra:
From the definition of $V$ and how integration works, the volume we need to calculate is $$\int_{0}^{\infty}\int_{0}^{\infty}\max\left(0,\min\left(2-x^{2}-y,x+2y\right)\right)\,\mathrm{d}x\,\mathrm{d}y$$
We know this works because the "$\max(0$" part ensures $z\ge0$ and that we do not accidentally include negative signed volume for points where the $\min$ is negative.
Resolving the min:
Now, we first need to resolve $\min\left(2-x^{2}-y,x+2y\right)$. The key will be to find the points where both expressions are equal, and then we can hopefully see which side of those points we're on. $2-x^{2}-y=x+2y$ simplifies by algebra to $y=\dfrac{2-x-x^{2}}{3}$. If $y$ increases, then $x+2y$ increases and $2-x^{2}-y$ decreases, so that $2-x^{2}-y$ would be the $\min$. And analogously if $y$ were less than that. Therefore, $$\min\left(2-x^{2}-y,x+2y\right)=\begin{cases} x+2y & \text{ if }y\le\dfrac{2-x-x^{2}}{3}\\ 2-x^{2}-y & \text{ if }y\ge\dfrac{2-x-x^{2}}{3} \end{cases}$$
This changes our integral into the following:
$$\int_{0}^{\infty}\int_{0}^{\infty}\max\left(0,\begin{cases} x+2y & \text{ if }y\le\dfrac{2-x-x^{2}}{3}\\ 2-x^{2}-y & \text{ if }y\ge\dfrac{2-x-x^{2}}{3} \end{cases}\right)\,\mathrm{d}x\,\mathrm{d}y$$
$$=\int_{0}^{\infty}\int_{0}^{\infty}\begin{cases} \max\left(0,x+2y\right) & \text{ if }y\le\dfrac{2-x-x^{2}}{3}\\ \max\left(0,2-x^{2}-y\right) & \text{ if }y\ge\dfrac{2-x-x^{2}}{3} \end{cases}\,\mathrm{d}x\,\mathrm{d}y$$
$$=\int_{0}^{\infty}\int_{0}^{\infty}\begin{cases} \max\left(0,x+2y\right) & \text{ if }y\le\dfrac{2-x-x^{2}}{3}\\ \max\left(0,2-x^{2}-y\right) & \text{ if }y\ge\dfrac{2-x-x^{2}}{3} \end{cases}\,\mathrm{d}y\,\mathrm{d}x$$
$$=\int_{0}^{\infty}\left(\int_{0}^{{\displaystyle \max\left(0,\dfrac{2-x-x^{2}}{3}\right)}}\max\left(0,x+2y\right)\,\mathrm{d}y\right.$$
$$\left.+\int_{{\displaystyle \max\left(0,\dfrac{2-x-x^{2}}{3}\right)}}^{\infty}\max\left(0,2-x^{2}-y\right)\,\mathrm{d}y\right)\,\mathrm{d}x$$
Resolving each max:
Now we need to resolve each $\max$.
The max in the bounds:
Let's start with the one in the bounds of the integrals. $\max\left(0,\dfrac{2-x-x^{2}}{3}\right)$ switches when the expressions are equal. $0=\dfrac{2-x-x^{2}}{3}\Rightarrow x^{2}+x-2=0$. And since $x^{2}+x-2=\left(x-1\right)\left(x+2\right)$, the $\max$ switches when $x=1$ or when $x=-2$. But since we know the outer integral only depends on positive values of $x$, we don't care about $x=-2$. When $0\le x\le1$, $\max\left(0,\dfrac{2-x-x^{2}}{3}\right)=\dfrac{2-x-x^{2}}{3}$ and when $x\ge1$, $\max\left(0,\dfrac{2-x-x^{2}}{3}\right)=0$.
Therefore, our integral above can be written as three iterated integrals in the following way:
$$\int_{0}^{1}\left(\int_{0}^{\dfrac{2-x-x^{2}}{3}}\max\left(0,x+2y\right)\,\mathrm{d}y+\int_{\dfrac{2-x-x^{2}}{3}}^{\infty}\max\left(0,2-x^{2}-y\right)\,\mathrm{d}y\right)\,\mathrm{d}x$$ $$+\int_{1}^{\infty}\int_{0}^{\infty}\max\left(0,2-x^{2}-y\right)\,\mathrm{d}y\,\mathrm{d}x$$
The simplest max:
Next, let's examine $\max\left(0,x+2y\right)$. This switches when $y=-\dfrac{x}{2}$, but for positive $x$ (as we have $\int_{0}^{1}\cdots\mathrm{d}x$) that's a negative value, and $\int_{0}^{\dfrac{2-x-x^{2}}{3}}\cdots\mathrm{d}y$ doesn't allow $y$ to be negative. So a switch can't happen, and we can test any value. For instance, if $x=\frac{1}{2}<1$ and $y=\frac{1}{4}<\dfrac{2-x-x^{2}}{3}$, then $x+2y$ is positive since it's the sum of two positive values (and it equals $1$). So for the purposes of our first iterated integral, $\max\left(0,x+2y\right)=x+2y$.
The final max:
Finally, let's examine $\max\left(0,2-x^{2}-y\right)$. This switches when $y=2-x^{2}$. For the second iterated integral, it matters how this compares to $\dfrac{2-x-x^{2}}{3}$ when $x\in\left[0,1\right]$. Well, $2-x^{2}=\dfrac{2-x-x^{2}}{3}$ when $x=\dfrac{1\pm\sqrt{33}}{4}$. The plus sign makes $x>1$ and the minus sign makes $x<0$, so by testing a point like $x=\frac{1}{2}$, we can be sure that $2-x^{2}>\dfrac{2-x-x^{2}}{3}$ always when we're within the bounds of the second iterated integral. Then for the integrand, we can test a point like $x=\frac{1}{2}<1$ and $y=\frac{1}{2}>\dfrac{2-x-x^{2}}{3}$; then $2-x^{2}-y=\frac{5}{4}>0$. So $\max\left(0,2-x^{2}-y\right)=2-x^{2}-y$ until $y=2-x^{2}$. For the third iterated integral, $2-x^{2}\ge0$ for $x\ge1$ whenever $x\le\sqrt{2}$; for any larger $x$ with $y$ nonnegative, $2-x^{2}-y$ is clearly negative.
The answer:
Putting this all together, we have the following:
$$\int_{0}^{1}\left(\int_{0}^{\dfrac{2-x-x^{2}}{3}}x+2y\,\mathrm{d}y+\int_{\dfrac{2-x-x^{2}}{3}}^{{\displaystyle 2-x^{2}}}2-x^{2}-y\,\mathrm{d}y\right)\,\mathrm{d}x$$ $$+\int_{1}^{\sqrt{2}}\int_{0}^{{\displaystyle 2-x^{2}}}2-x^{2}-y\,\mathrm{d}y\,\mathrm{d}x$$