S={(x ,y ,z) ЄR³ : x² + y² ≤ (2-z)² , 0 ≤ z ≤ 1}
with variable density(x, y, z)= -ln(z)/√x² + y²
I have no clue how to do this...please help...
Surely you know that the mass of an object is its density integrated over its volume.
Your volume is bounded by the cone $\displaystyle x^2+ y^2= (2- z)^2$, below by z= 0 and above by z= 1. The only thing peculiar about this is that the vertex of the cone is at z= 2. Since z<1 2- z is below 2 and we can use $\displaystyle z= 2- \sqrt{x^2+ y^2}$. Projecting onto the xy-plane, we get circle with center at (0,0) and radii 1 to 2.
In x, y, z coordinates, we would have to integrate over the 'washer' between the circle with radius 1 and the circle with radius 2. That can be done but would require four separate integrals (from x= -2 to x= -1, from x= -1 to 1 with y above the smaller circle, from x= -1 to 1 with y below the smaller circle, from x= 1 to 2). It is much easier to use cylindrical coordinates. In that case, r goes from 1 to 2, [itex]\theta[/itex] goes from 0 to 2\pi, and, for each r and [itex]\theta[/itex], z goes from the plane, z= 1 up to the cone, [tex]z= 2-\sqrt{x^2+ y^2}= 2- \sqrt{r^2}= 2- r.
In cylindrical coordinates the density function is $\displaystyle \frac{-ln(z)}{\sqrt{x^2+ y^2}}= \frac{-ln(z)}{\sqrt{r^2}}= \frac{-ln(z)}{r}$.
That is your integral is $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 1}^2\int_{z=1}^{2- r} \frac{-ln(z)}{r}(r drd\theta dz)= -2\pi\int_{r= 1}^{2}\int_{z=1}^{2-r} ln(z)dzdr$.
Cylindrical coordinates would simplify things a good deal. The boundary of the cone, in cylindrical form, is
$\displaystyle r^2 = (2-z)^2$
$\displaystyle \Rightarrow r = |2-z| = (2 - z)$ (since $\displaystyle 0\leq z\leq1$ and we want $\displaystyle r\geq0$)
So we have
$\displaystyle 0\leq r\leq 2-z$
$\displaystyle 0\leq\theta\leq2\pi$
and
$\displaystyle 0\leq z\leq 1.$
Now, we integrate the density function over this region to find the mass:
$\displaystyle m = \iiint\limits_S\frac{-\ln z}{\sqrt{x^2+y^2}}\,dV$
$\displaystyle =\int_0^1\int_0^{2\pi}\int_0^{2-z}\frac{-r\ln z}r\,dr\,d\theta\,dz$
$\displaystyle =-\int_0^1\int_0^{2\pi}\int_0^{2-z}\ln z\,dr\,d\theta\,dz$
Thank you guys so much!!!! I really appreciate your help. I have another question!
How should I go on now? Cuz if I start to integrate I have problem at ln(0) what is not defined! How should I handle this now!
I know I have a lot of stupid questions, but I really try to pass that math class...:-(
Just in case a picture helps work the integral from the inside out...
... where (key in spoiler) ...
Spoiler:
__________________________________________________ __________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
Right, I didn't notice that. This will actually be an improper integral:
$\displaystyle =-\int_0^1\int_0^{2\pi}\int_0^{2-z}\ln z\,dr\,d\theta\,dz$
$\displaystyle =-2\pi\int_0^1(2-z)\ln z\,dz$
$\displaystyle =-2\pi\lim_{a\to0^+}\int_a^1(2-z)\ln z\,dz$
$\displaystyle =-2\pi\lim_{a\to0^+}\left[\left(2z-\frac12z^2\right)^2\ln z+\frac14z^2-2z\right]_a^1$ (See tom's post)
$\displaystyle =-2\pi\lim_{a\to0^+}\left[0 + \frac14 - 2 - \left(2a-\frac12a^2\right)\ln a - \frac14a^2 + 2a\right]$
$\displaystyle =-2\pi\lim_{a\to0^+}\left[-\frac74+\left(2a-\frac12a^2\right)\ln a - \frac14a^2 + 2a\right]$
$\displaystyle =-2\pi\left(-\frac74\right) - 2\pi\lim_{a\to0^+}\left(2a-\frac12a^2\right)\ln a$
$\displaystyle =\frac{7\pi}2-\pi\lim_{a\to0^+}\left(4a-a^2\right)\ln a$
$\displaystyle =\frac{7\pi}2-\pi\lim_{a\to0^+}\frac{\ln a}{1/\left(4a-a^2\right)}$
This limit produces the indeterminant form $\displaystyle \textstyle-\frac\infty\infty,$ so we can apply L'Hôpital's rule,
$\displaystyle =\frac{7\pi}2-\pi\lim_{a\to0^+}\frac{1/a}{-(4-2a)/\left(4a-a^2\right)^2}$
$\displaystyle =\frac{7\pi}2+\pi\lim_{a\to0^+}\frac{\left(4a-a^2\right)^2}{a(4-2a)}$
$\displaystyle =\frac{7\pi}2+\pi\lim_{a\to0^+}\frac{a\left(4-a\right)^2}{4-2a}$
$\displaystyle =\frac{7\pi}2.$