# Limits of integration on double integrals

I was given this problem:

Find an integral equal to the volume of the solid bounded by $$z=4-2y,z=0,x=y^4,x=1$$ and evaluate.

I understand how to evaluate once my double integral is set up, but I do not know how to find my limits of integration.
I am assuming that my function will be $$z=4-2y$$ and that using this I should be able to find my limits of integration. I can say that $$0=4-2y$$ which means that $$y=2$$. I can then plug that into $$x=y^4$$ and get $$1\leq x\leq 16$$ which may be correct, but I still am missing the limits of integration for y.

• Draw a picture in the $z=0$ plane. – saulspatz Apr 28 at 16:34

Note that $$x=y^4$$ and $$x=1$$ intersect at $$(x,y)=(1,\pm1)$$. which define the limits for the integration region in the $$xy$$- plane. Thus, the volume integral is

$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$

That solid is located above the plane $$z=0$$ and below the plane $$z=4-2y$$. The possible values for $$x$$ belong to the $$[0,1]$$ interval (the condition $$x=y^4$$ prevents $$x$$ from being negative). So, you should compute$$\int_0^1\int_{-\sqrtx}^{\sqrtx}\int_0^{4-2y}1\,\mathrm dz\,\mathrm dy\,\mathrm dx\tag1$$But\begin{align}(1)&=\int_0^1\int_{-\sqrtx}^{\sqrtx}4-2y\,\mathrm dy\,\mathrm dx\\&=\int_0^18\sqrtx\,\mathrm dx\\&=\frac{32}5.\end{align}

• Interesting - this is a slightly different method that still works! Thank you. 4 – Burt Apr 29 at 17:06
• As I'm further understanding all these methods, I don't understand why x needs to be from zero to one. Why can't it be from one and higher? – Burt Apr 29 at 21:28
• Your solid is bounded by the plane $x=1$, right?! – José Carlos Santos Apr 29 at 21:47

As you know how to evaluate the integral - and as it has been evaluated in other answers! - I'll concentrate on showing that a unique subset of $$\mathbb{R}^3$$ is bounded by the four surfaces identified in the question, and describing that subset in such terms that one can write down the triple integral that is to be evaluated.

@saulspatz's comment recommends first drawing a figure, ignoring the $$z$$ coordinate. I also find this to be the easiest way to think about the question.

The plane $$x = 1$$ cuts the $$(x, y)$$ plane in a line, and the surface $$x = y^4$$ cuts the $$(x, y)$$ plane in a curve. The line and curve together subdivide the $$(x, y)$$ plane into five subsets, which correspond to four subsets of $$\mathbb{R}^3$$: \begin{align*} A & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \geqslant 1 \}, \\ B & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \leqslant 1 \}, \\ C & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \geqslant 1 \}, \\ D & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \leqslant 1 \}. \end{align*} Each of $$A, B, C, D$$ is a connected subset of $$\mathbb{R}^3,$$ but the projection of $$C$$ on the $$(x, y)$$ plane has two separate components, corresponding to positive and negative values of $$y.$$

Each of $$A, B, C, D$$ is an unbounded subset of $$\mathbb{R}^3,$$ but the projection of $$D$$ on the $$(x, y)$$ plane is bounded. That looks hopeful! In order to be in a position to say something more definite than that, the easiest thing to do next (or so I think) is to look at the projection of the planes $$z = 4 - 2y$$ and $$z = 0$$ on the $$(y, z)$$ plane.

One can see that these two planes between them divide $$\mathbb{R}^3$$ into four subsets: \begin{align*} E & = \{ (x, y, z) \colon (y \geqslant 2 \text{ and } z \geqslant 0) \text{ or } (y \leqslant 2 \text{ and } z \geqslant 4 - 2y) \}, \\ F & = \{ (x, y, z) \colon (y \leqslant 2 \text{ and } z \leqslant 0) \text{ or } (y \geqslant 2 \text{ and } z \leqslant 4 - 2y) \}, \end{align*} \begin{align*} G & = \{ (x, y, z) \colon 4 - 2y \leqslant z \leqslant 0 \}, \\ H & = \{ (x, y, z) \colon 0 \leqslant z \leqslant 4 - 2y \}. \end{align*}

Subset $$E$$ contains points with arbitrarily large positive values of $$z$$ for any value of $$y$$; and subset $$F$$ contains points with arbitrarily large negative values of $$z$$ for any value of $$y$$; therefore neither $$E$$ nor $$F$$ has a bounded intersection with any of $$A, B, C, D.$$

Subset $$G$$ only contains points with values of $$y \geqslant 2,$$ therefore its intersection with $$D$$ is empty.

Subsets $$A, B, C$$ all have points with arbitrarily large positive values of $$y,$$ as do their intersections with $$G.$$

Therefore the only candidate for a subset of $$\mathbb{R}^3$$ that is bounded by the four given surfaces - and is bounded (!) - is: $$D \cap H = \{ (x, y, z) \colon y^4 \leqslant x \leqslant 1 \text{ and } 0 \leqslant z \leqslant 4 - 2y \}.$$ This is indeed bounded, and we can evaluate the volume integral by writing: $$\int_{D \cap H} 1 = \int_{-1}^1\int_{y^4}^1\int_0^{4 - 2y}\,dz\,dx\,dy.$$ I'll stop here - approximately where the other answers start. :)

• Thank you for this detailed answer! I can just choose to ignore the z plane, right? – Burt Apr 29 at 17:04
• The $x, y$ and $z$ coordinates are all involved, but (as in @saulspatz's comment) I found it easiest to ignore the $z$ coordinate to begin with, i.e. draw a "plan", ignoring the "elevation" for the moment. This allows you to form a clear view of the figure formed by the $x = y^4$ and $x = 1$ surfaces. You can then consider the two remaining surfaces $z = 4 - 2y$ and $z = 0.$ Again I found it helpful to look at these two separately from the other two at first, even though one can already begin to see how they will intersect the first pair of surfaces. My aim was to avoid all guesswork. – Calum Gilhooley Apr 29 at 17:25
• More succinctly: ignore $z$ (in the sense of ignoring the surfaces whose equations involve $z$), then ignore $x$ (in the sense of ignoring the surfaces whose equations involve $x$), then put the two views together. – Calum Gilhooley Apr 29 at 18:58