Why this simple volume problem in Multivariate Calculus seems to have an anomaly?

Question

Find the volume of the solid inside the cylinder $x^2+y^2-2ay = 0$ and between the plane $z = 0$ and the cone $x^2+y^2 = z^2$.

I tried solving this problem as follows:

Equation of the cylinder $x^2+(y-a)^2 = a^2$. Now, if we are able to find the volume $V$ between the plane $z = 0$ and $z = \sqrt{x^2+y^2}$, then the required volume is $2V$.

Let $E = \{ (x,y,z) \in \Bbb R^3: (x,y) \in D \text{ and } 0 \leq z \leq \sqrt{x^2+y^2} \}$ , where $D$ is the disk $x^2+(y-a)^2 = a^2$.

To find $V$ we note that, $V = \iiint_EdV=\iint_D\int_{0}^{\sqrt{x^2+y^2}}dzdA = \iint_D\sqrt{x^2+y^2}dA$.

Converting to polar coordinates we have, $V = \int_0^{2\pi}\int_0^{2a\sin\theta}r^2drd\theta = 0$.

So, the total volume is $2V=0$.

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Now, this is clearly wrong. However, if we evaluate the same integral as follows:

$V = 2\int_0^{\pi}\int_0^{2a\sin\theta}r^2drd\theta$.

Then $2V$ gives the required volume.

But $D$ consists of the whole circle and so taking the limits $[0,2\pi]$ and multiplying with $2$ or $[0,\pi]$ and multiplying by $4$ shouldn't make any difference. I want to know the reason behind this strange anomaly.

Answer 1

$D$ is a disk, but not a disk centered at the origin. In fact, $y\geq0$ in $D$, so you need $\sin\theta\geq0$, which happens when $\theta\in[0,\pi]$. That is, $$ D=\{(r,\theta):\ \theta\in[0,\pi],\ 0\leq r\leq2a\sin\theta\}. $$

Answer 2

Basically , when we use change of variables , the new variables must be monotonically increasing to cover the range.

Here , $\sin \theta$ is not monotonically increasing over $(0,2\pi)$. That will mess up the Integral , adding & subtracting values here & there.

The Best approach would be to consider 4 Integrals for 4 Quadrants. In each Quadrant , $\sin \theta$ is monotonically increasing (maybe with reversal of the limits)
Add up these 4 Integral values to get the whole.

In your Case , these are getting added & subtracted to give $0$ value !
When adding the 4 Quadrants , we will get the correct value.

UPDATE :

I realize that there are multiple nuances not covered in this Short Post.
I also realize that change of variables need not be monotonic which user Eric Towers rightly mentioned in the Comments.

I do think that when it is not monotonic , there are complications to handle ; Other-wise we will get wrong Integral values , like the $0$ in the Current Post.

Here is my Example to high-light that.
Say we want to Integrate $x+y^2$ over this Purple Curve :

Let us assume we are given the Curve in PBAL [ Parameterized By Arc Length ] terms. At Origin , Arc length $t=0.0$ & at right Blue Inter-Section , $t=0.7$ , which are the limits we want to Integrate over.

Now , we use change of variables & convert $s$ to $y$ & we see that the new limits are $y=0$ & $y=6$.
When we Integrate , we might unfortunately get only the Section from Origin ( $t=0$ ) to the left Blue Intersection ( $t=0.3$ ) , because $y$ was not monotonic with $t$.

In Case , we had taken the Integral from $y=0$ to $y=8$ & then added up the Integral from $y=8$ to $y=6$ ( while handling the Directions too ! ) then we would have got the right Integral value.

In Case , we had taken change of variables with monotonicity , the Integral will work out automatically.

In Summary , I agree with Eric Towers , that monotonicity is not necessary , though I do think that we have to be careful when monotonicity is not available.