Unleash Your Finding Volume Potential with Expert Support and Comprehensive Practice

Probability of finding $n_1$ particles in volume $v_1$ and $n_2$ particles in volume $v_2$.
The formula is:
$P[n\text{ particles in }V]=(\genfrac{}{}{0ex}{}{N}{n})p^n(1-<!--\; -\; -->p{)}^{N-<!--\; -\; -->n}$ where p is $p=\frac{v}{V}$.
I would like to know how this formula generalizes if one wants to compute "the probability of finding $n_1$ particles in volume $v_1$ and $n_2$ particles in volume $v_2$, given that there are N indistinguishable particles in total, that the total volume is V and that $v_1$ and $v_2$ are disjoint."

Finding volume of a cone given density
Let C be the solid cone with the boundary surfaces $x^2+y^2=z^2$ and $z=0$. The density of the solid at point (x,y,z) is z.
Find the volume of the solid using the integrals in both the cylindrical coordinates and the spherical coordinates.
I really cant do this question. I know that $V=\mathrm{\rho <!--\; \rho \; -->}/m$ but what is m?
I also don't understand how you can even find the volume with integrals because there is no limit of z given so it would just be infinity wouldn't it.
Can someone start me off, I obviously don't need anyone to show me how to integrate. Just struggling on what the density and stuff is relevant.

Volume of trig function around y-axis
I have this question and it's the first kind of question I'm doing involving finding volume so I just would like some help solving this question:
Find the volume created by revolving the curve $y=\frac{\sin <!--\; \; -->(x)}{x}$ from $x=\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$ to $x=\mathrm{\pi <!--\; \pi \; -->}$. about the y-axis

Find the volume of a tetrahedron with vertices:
O(0,0,0), A(1,2,3), B(-2,1,5), C(3,7,1) by using triple integral
First find the the equations of the planes.

Volume between a sphere and cone
I'm having some problems finding the volume between a sphere of radius 5 and the cone $z=-<!--\; -\; -->\sqrt{x^2+y^2}$.
The bounds I got for spherical coordinates are 0 to $2\mathrm{\pi <!--\; \pi \; -->}$ for $\mathrm{\theta <!--\; \theta \; -->}$, $3\mathrm{\pi <!--\; \pi \; -->}/4$ to $\mathrm{\pi <!--\; \pi \; -->}$ for $\mathrm{\varphi <!--\; \varphi \; -->}$, 0 to 5 for r, however I'm not sure if these are correct. My thought process is that if you let say x or $y=0$ in the equation for a cone, you get an inverted modulus graph which is where I got the bounds for $\mathrm{\varphi <!--\; \varphi \; -->}$.
For cylindrical coordinates I got bounds of 0 to $2\mathrm{\pi <!--\; \pi \; -->}$ for $\mathrm{\theta <!--\; \theta \; -->}$, 0 to 5 for r and 0 to $\sqrt{25-<!--\; -\; -->r^2}$ for z and after evaluating the triple integral I got a volume of 125/3.
Could someone run me through their thought process and show me where I went wrong?

Find the volume of intersection between cylinders
$x^2+y^2\le <!--\; \le \; -->1$, $x^2+z^2\le <!--\; \le \; -->1$, $y^2+z^2\le <!--\; \le \; -->1$.
I am having tough time finding the volume how do i solve this kind of questions?
My trial : i will move to the cylinder coordinates of the xy cylinder let:
$x^2+y^2=r^2$
$z=z$
$0\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$
Solving the inequalties i get:
$0\le <!--\; \le \; -->r^2\le <!--\; \le \; -->1$
$-<!--\; -\; -->\sqrt{1-<!--\; -\; -->\frac{r^2}{2}}\le <!--\; \le \; -->z\le <!--\; \le \; -->\sqrt{1-<!--\; -\; -->\frac{r^2}{2}}$
$0\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$
The integral is:
${\int <!--\; \int \; -->}_{z=-<!--\; -\; -->\sqrt{1-<!--\; -\; -->\frac{r^2}{2}}}^{z=\sqrt{1-<!--\; -\; -->\frac{r^2}{2}}}{\int <!--\; \int \; -->}_{r=0}^{r=1}{\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}dzdrd\mathrm{\theta <!--\; \theta \; -->}$ = $\frac{4\mathrm{\pi <!--\; \pi \; -->}}{\sqrt{2}}\frac{(2-<!--\; -\; -->r^2{)}^{\frac{3}{2}}}{-<!--\; -\; -->3}{|}_{r=0}^{r=1}$

Finding the volume when the region between $y=x^2$ and $y=4-<!--\; -\; -->x^2$ is rotated about the x-axis.
So I start by finding the roots where they meet, so I find:
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\sqrt{2}}^{\sqrt{2}}\mathrm{\pi <!--\; \pi \; -->}(4-<!--\; -\; -->x^2{)}^{2}dx$
But this got me the wrong answer 75.825. Is that right or is the book wrong with 94.7?

Finding the volume of two subsets of ${\mathbb{R}}^3$.
Define two subsets of ${\mathbb{R}}^3$.
$A=\{(x,y,z)\in <!--\; \in \; -->{\mathbb{R}}^3:|x|+|y|+|z|\le <!--\; \le \; -->1\}$
$B=\{(x,y,z)\in <!--\; \in \; -->{\mathbb{R}}^3:max\{|x|,|y|,|z|\}\le <!--\; \le \; -->1\}$
1. Find vol(B)-vol(A)
2. Set up an integral which evaluates to vol(A)-vol(B)

Area below x axis not considered when finding volume of shape rotated around y axis?
While practicing finding the volume of shapes bounded by a function and the x-axis by rotating the shape around the y-axis and taking an integral (the "washer" or "pipe" method) I became confused as to why the shape used to calculate volume "stops" at the x axis, and isn't a part of the calculation.
To explain further, if the shape is being discretized in the x direction (dx), then the integral will be set up to be calculated between two x values, call them a and b. This restricts the calculation to be determining the area only between the x values of a and b. In the y direction, however, there is no such restriction, and so far I have been assuming that the y component of the calculation "begin" at the function being integrated and "end" at the x-axis. I do not understand, however, the mathematics behind why I get the correct answers for the volume of the rotated shape if I am ignoring the area that extends from the x axis to negative infinity in the y direction.
A simple example that can help illustrate my question is finding the volume of the shape defined by $y=x^2$ between $0<x<2$ rotated around the y axis. Why, in this example, is the area below the x axis not considered?

Integrating over or under this object
There is clearly a kink in my understanding of double integrals with polar coordinates.
Our problem is to find the volume enclosed by the hyperboloid $-<!--\; -\; -->x^2-<!--\; -\; -->y^2+z^2=1$ and plane $z=2$. I correctly figure the bounds to be $D=\{(r,\mathrm{\theta <!--\; \theta \; -->})|0\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->},0\le <!--\; \le \; -->r\le <!--\; \le \; -->\sqrt{3}\}.$
But I have trouble finding if we are integrating between the section of the hyperboloid and the disk $r=1$ at $z=0$ or $z=2$.
While $r\sqrt{1-<!--\; -\; -->r^2}drd\mathrm{\theta <!--\; \theta \; -->}$ would make some sense as an integrand, I essentially dont know if I am integrating (ie, finding volume) under or over the surface of the hyperboloid.

Volume of the intersection of two lp balls.
Given (1) a d-dimensional space,
(2) a $l_p$ ball of radius $r_1$, and
(3) a $l_q$ ball of radius $r_2$, where $0<p<q\le <!--\; \le \; -->2$,
(4) both balls are centered on the origin.
Please can someone help me in finding the volume of the intersection of these two balls?
In low dimensional space, for example $d<10$, I can compute the volumn by applying monte carlo simulation. However, the monte carlo method does not work in high dimensional space, e.g. d > 100, because the number of samples required is huge, which is beyond the computational power. Therefore, I wish to get the formula of the volume with respect to d, p, q, $r_1$ and $r_2$.

Finding a Function for Volume
A carpenter has been asked to build an open box with a square base. The sides of the box will cost 4 dollars per square metre and the base will cost 5 dollars per square metre. 60 dollars is available to build the box. If the dimensions of the box are x, the side of the base, and y, the height of the box, find a function for the volume in terems of one of the variables.

Calculus Disk/Washer Method for Volume
I am given the bounded functions $y=ln(x),g(x)=-<!--\; -\; -->.5x+3$, and the x-axis. The reigon R is bounded between these, and I'm tasked with finding the volume of this solid using disk/washer method when revolved around the x-axis.
I know the formula I need to use, but I'm a little confused on finding the upper and lower limits and which to place as an innner and outer radius since the functions aren't graphed like the traditional washer method problem.

Question about finding volume using integration
I need to solve the volume that's between:
$z=0,x=0,y=0,x^2+y^2=4z=12-<!--\; -\; -->x-<!--\; -\; -->y$
1) Does it matter if I use double integrals or triple?
2) When I draw this area on the x and y plane, I have a circle with radius 2. How do I know which quadrant my volume is in?

Finding volume of region bound by the curves $y={\sin }^2<!--\; \; -->x$, $y={\cos }^2<!--\; \; -->x$ and he y-axis about $x=2$.
Find the volume of the solid generated by revolving the region bounded by the curves $y={\sin }^2<!--\; \; -->x$, $y={\cos }^2<!--\; \; -->x$ ($0\le x\le \frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$ and the y-axis about $x=2$ using both the disk/washer and cylindrical shell methods.
I know how to use the above method if it is a revolution around the axes but how do we use it to find the volume around a line?

Finding volume of a revolution
I want to find the volume of the revolution that occurs when the region bounded by $y=x^2$ and $y=1$ is revolved around the line $y=2$. The problem is that it is not solid and I cannot understand how the disk method could be used. How can one attack the problem?

Finding volume by rotating cos x and sin x about $y=-<!--\; -\; -->1$, $x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$
I'm working on a problem that is asking for the volume of $y=\cos <!--\; \; -->x$ when rotated about the line $y=1$, with a restricted domain of $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. The range is [0, 2].
There's a problem I'm working on that is asking for the volume of $y=\cos <!--\; \; -->x$ and $y=\sin <!--\; \; -->x$ when rotated about line $y=-<!--\; -\; -->1$ with a domain of $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$
I don't even know where to begin. First off; which method would be most effective in this case? A shell, washer or disc?

Finding the volume of the solid generated by revolving the given curve.
The objective is to find the volume of the solid generated by revolving the curve $y={\displaystyle \frac{a^3}{a^2+x^2}}$ about its asymptote.
Observing the given function yields that $y\ne <!--\; \ne \; -->0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-<!--\; -\; -->axis$ is gives as $V=2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}(f(x{)}^2-<!--\; -\; -->0)dx$.
Which gives: $2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{\displaystyle \frac{a^6}{(a^2+x^2{)}^2}}dx$
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form ${\displaystyle \frac{1}{x^4+2x^2+1}}$ for $a=1$, which is transformed into ${\displaystyle \frac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}}=2[{\displaystyle \frac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}}-<!--\; -\; -->{\displaystyle \frac{(1-<!--\; -\; -->\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}}]$ which can be integrated, but this integral is tending to infinity.

Finding volume inside a paraboloid and inside a sphere
I want to find the volume inside the paraboloid $z=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=12$.
So, I wrote $x=r\cos <!--\; \; -->t,y=r\sin <!--\; \; -->t$, find out $0\le <!--\; \le \; -->r\le <!--\; \le \; -->\sqrt{3}$ and $0\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$. We also have $z=r^2$ the paraboloid and $z=\sqrt{12-<!--\; -\; -->r^2}$. Now, why is the desired volume is given as:
${\displaystyle {\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^{\sqrt{3}}(\sqrt{12-<!--\; -\; -->r^2}-<!--\; -\; -->r^2)rdrdt?}$
Why do we subtract two z values in this way and not use just one of them?

Finding volume using disks/washers
Find the volume of a solid obtained by rotating the region bounded by $y=\sqrt{x}$ and $y=x$ about $y=2$.