Why is this domain wrong?

Question

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:

$$D=\{(\theta,r,x):0<\theta<2\pi,0<r<\sqrt{\frac x4},0<x<4\} $$

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:

$$B=\{(\theta,r,x):0<\theta<2\pi,0<r<1,0<x<4r^2\}$$

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 $B$ & $D$ ?

Answer 1

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

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

Answer 2

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.