# Why is this domain wrong?

I was trying to integrate:

$$\iiint_D x\ dx\,dy\,dz$$

where $$D$$ is limited by $$x=4y^2+4z^2$$ and $$x=4$$.

I have managed to find the answer by converting the domain to cylindrical coordinates, using:

$$z = r \sin(\theta)$$
$$y = r \cos(\theta)$$
$$x = x$$

and the domain:

$$D=\{(\theta,r,x):0<\theta<2\pi,0

Integrating over this domain, I got the right answer which is:

$$\iiint_D x\ dx\,dy\,dz=\frac {16\pi}{3}$$

However, this was not my first choice for a domain. I first tried:

$$B=\{(\theta,r,x):0<\theta<2\pi,0

Which, when evaluated in the integral, gives:

$$\iiint_B x\ dx\,dy\,dz=\frac {8\pi}{3}$$

I understand that since the results are different, these domains are not equivalent. Moreover, since I know the right answer for this problem, I know that the second domain is wrong. But I can not understand why. Solving the integral is relatively easy, but finding the domain is not always very clear to me.

Why is the first domain $$D$$ right and the second domain $$B$$ is wrong in this case?

What is the Connection between $$B$$ & $$D$$ ?

• @Cironis your comment is not appropriate. Commented Sep 27, 2023 at 18:29
• @Cironis What a horrifyingly elitist comment. Is it hard to comprehend that people find different things difficult? You've answered no questions on the site, so what are you actually trying to do here?
– kipf
Commented Sep 27, 2023 at 18:29
• That is ok. I am used to this kind of treatment to my questions. There is always a good soul that helps me in the end.. Commented Sep 27, 2023 at 18:31
• To @user3347814: $D$ is limited by $x=4y^2+4z^2$ and $x=4$? Commented Sep 27, 2023 at 18:35
• The surface $x=4y^4+4z^4$ in polar form is not equal to $x=4r^2$. Also, $x$ is bounded below by $4y^4+4z^4$ and above by $4$, so your inequality should be $$4y^4+4z^4<x<4$$
– user801306
Commented Sep 27, 2023 at 18:53

Assuming typo that the exponents are 2 on $$y$$ and $$z$$ not $$4$$. The rest of your question uses that as the bounds.

The $$\theta$$ variable is irrelevant so let's just fix it to $$0$$ in both $$D$$ and $$B$$.

For $$D$$ we have $$0 \lt r \lt \sqrt{\frac{x}{4}}$$ and $$0 \lt x \lt 4$$, so we can consider $$x=1$$, $$0 < r \lt \frac{1}{2}$$.

Meanwhile, lets fix the same value of $$x$$ in $$B$$, we get the constraint $$4r^2 \gt 1$$, which means $$1 \gt r \gt \frac{1}{2}$$ when combined with the $$0 \lt r \lt 1$$.

It's the wrong way around. D and B are (interiors of) complements of each other inside the cylinder of height 4.

Add up the answers you get for D and B and you get $$\frac{24\pi}{3} = 8 \pi$$. Let $$D+B$$ be that cylinder

$$I_{D+B} \equiv \int_0^{2\pi} \int_0^1 \int_0^4 x \; r \; dx \; dr \; d\theta\\ = 2 \pi \int_0^1 \int_0^4 x \; r \; dx \; dr\\ = 2 \pi \int_0^1 r \; 8 \; dr\\ = 16 \pi \int_0^1 r dr\\ = 8 \pi\\ I_{D+B} = I_D + I_B$$

• +1 , this is the right thinking. I was going to Post this , but you got it earlier. I might Post my own Answer , with a little more Detail ....
– Prem
Commented Sep 27, 2023 at 19:18

The Answer by user AHusain is right , OP is going to get the Complements.

Here , I will give a visual way to see that Issue & then give the way to get it right.

In this Image , I am using arbitrary Curve , which is the Straight line in OP Case.

Let the $$Z$$ Dimension be ignored here. It will come in , when $$\theta$$ varies. When $$\theta=0$$ , we will get the Area in the $$XY$$ Plane.
In the $$XY$$ Plane , $$y$$ & $$r$$ are Same.

(A) Original Order : When we let $$x$$ vary between $$0$$ & $$4$$ , then let $$y=r$$ vary between $$0$$ & $$\sqrt{x/4}$$ , we will rightly get the Area below the curve in the $$XY$$ Plane , where we are using the vertical green lines.

(B) Changing the order Wrongly : When we let $$y=r$$ vary between $$0$$ & $$1$$ , then let $$x$$ vary between $$0$$ & $$4y^2=4r^2$$ , we are using the horizontal green Solid lines in the Image.
That is the Area above the Curve !
In other words , "(B) + (A) = total rectangle" in $$XY$$ Plane.
In general , that "total rectangle" will be between the bottom-left corner limits & the top-right corner limits.

(C) Changing the order Correctly : When we let $$y=r$$ vary between $$0$$ & $$1$$ , then let $$x$$ vary between $$4y^2=4r^2$$ & $$1$$ , we are using the horizontal green Dotted lines in the Image.
That is the Area below the Curve.
In other words , "(C) = (A)" in $$XY$$ Plane.
When we Integrate with those Correct limits , we will get Consistent Answer.

This Analysis is valid for arbitrary Curve , though OP Case has the Straight line.