Calculate $\iiint_Dz\;dxdydz$ if $D$ is the region inside $z=0,z=\sqrt{x^2+y^2}$ and $x^2+y^2=1$.
I would like to know if the answer I got is right. This is what I did:
$(1)$ Change to cylindrical coordinates
$$x=r\cos\phi,y=r\sin\phi,z=z$$
$(2)$ Determine the region in terms of $r,\phi,z$
I usually have some problems determining the region, let's see if I got this right:
$(2.1)$ From $z=0$ and $z=\sqrt{x^2+y^2}$ follows $0\leq z\leq \sqrt{r^2\cos^2\phi+r^2\sin^2\phi}=r$.
$(2.2)$ The regions goes the "full circle", so $0\leq\phi\leq 2\pi$.
$(2.3)$ The region is inside the unit circle, so $0\leq r\leq 1$.
$(3)$ Integrandus calculandus
$$\iiint_Dz\;dxdydz=\int_0^1\int_0^{2\pi}\int_0^r zr\;dzd\phi dr\\=\frac{1}{2}\int_0^1\int_0^{2\pi}r^3d\phi dr=\pi\int_0^1r^3dr=\frac{\pi}{4}.$$
Does it look right?.
I thought of an alternative way to "solve it" with cylindrical coordinates, I know I got the wrong answer but I cannot see why is this wrong.
Instead of the limits given before, I take
$$0\leq z \leq 1\\ 0\leq r \leq \sqrt{z}\\ 0\leq \phi\leq 2\pi $$
It seems that this describes the region inside the cone, and $$\int_0^1\int_0^{2\pi}\int_0^{\sqrt{z}}zr\;dr d\phi dz = \frac{1}{2}\int_0^1\int_0^{2\pi}z^2d\phi dz =\pi \int_0^1 z^2 = \frac{\pi}{3}.$$
The cone is inside the cilinder of raduis 1 from $0\leq z\leq 1$, then wouldn't be
$$\iiint_D = \text{volume cylinder – volume cone} = \pi – \pi/3 = 2\pi/3?$$
I know the answer is wrong (I already saw Danieltatis's answer) but I wanted to try a different solution, and I don't know where did I go wrong with the last one.
Best Answer
You were right, i am really sorry for confusing you without reading correctly.
The answer is $\pi/4$, because the proposed Integral
$\displaystyle\iiint\limits_Dz\;\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^1\int_0^{2\pi}\int_0^r zr\,\mathrm dz\,\mathrm d\phi\,\mathrm dr$
does cover the volume limited by the $45°$ cone, the $xy$ plane and the circular cylinder with axis over the $z$-axis and radius 1.