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Finding the volume of a solid enclosed by two cylinders and planes
Consider the solid E1 that is enclosed by the planes $z=0$ and $z=5$ and by the cylinders $X^2+y^2=9$ and $x^2+y^2=16$. Find the volume of E1. How to find the volume using two cylinders

Finding the volume of a parallel piped
Given the points $P(-<!--\; -\; -->2,3,1),Q(2,-<!--\; -\; -->2,0),R(4,1,0)$.
Find the parallelepiped with edge OP,OQ,OR where O is the origin (0,0,0).
Volume for a parallelepiped is $V=A\times <!--\; \times \; -->B\cdot <!--\; \cdot \; -->C$.
So would the volume be $OP\cdot <!--\; \cdot \; -->(OQ\times <!--\; \times \; -->OR)$.
$<-<!--\; -\; -->2,3,1>\cdot <!--\; \cdot \; -->(<2,-<!--\; -\; -->2,0>\times <!--\; \times \; --><4,1,0>)$
$<2,3,1>\cdot <!--\; \cdot \; --><0,0,10>$
area parallelepiped is 10 but would this be correct?

Finding the volume of a solid bounded by a sphere and a paraboloid
I am working on a problem that requires me to find the volume of the solid bounded by the sphere $x^2+y^2+z^2=2$ and the paraboloid $x^2+y^2=z$. I know that to do this, I must use triple integration.
Thus far, I know that we need to reevaluate the given equations and put them into spherical coordinates in order to achieve our bounds. When we've gotten our bounds, it's just simple triple integration to computer the volume.
So, for our sphere, I said that $x^2+y^2+z^2=2$ is equivalent to $\mathrm{\rho <!--\; \rho \; -->}=sqrt(2)$.
Then, for our sphere, I get a little messed up. I'm trying to find $\mathrm{\varphi <!--\; \varphi \; -->}$ boundaries here, but I get the equation ${\mathrm{\rho <!--\; \rho \; -->}}^{2}si{n}^2\mathrm{\varphi <!--\; \varphi \; -->}=\mathrm{\rho <!--\; \rho \; -->}cos\mathrm{\varphi <!--\; \varphi \; -->}$. Then, the $\mathrm{\rho <!--\; \rho \; -->}$ can cancel out, making it $\mathrm{\rho <!--\; \rho \; -->}si{n}^2\mathrm{\varphi <!--\; \varphi \; -->}=cos\mathrm{\varphi <!--\; \varphi \; -->}$, but my issue is that the $\mathrm{\rho <!--\; \rho \; -->}$ doesn't cancel out completely so I can't get the bounds for $\mathrm{\varphi <!--\; \varphi \; -->}$.

Finding volume of hyperboloid bounded by two planes
I want to find the volume bounded by hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-<!--\; -\; -->\frac{z^2}{c^2}=1$ and the planes $z=-<!--\; -\; -->c,z=c$.
I do not know whether should I use the cylindirical coordinates or spherical coordinates.
At first, I am thinking to set $x=au,y=bv,z=cw$ and now we have $u^2+v^2-<!--\; -\; -->w^2=1$. Jacobian of this transformation is abc. I guess that after this change of variables If I take a cross section then it will give me a circle on uv-plane but I do not know the reason.
If we consider the part of the volume that is inside the first octant, my intuition says that the desired volume is ${\displaystyle {\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->v^2}{\int <!--\; \int \; -->}_0^{u^2+v^2-<!--\; -\; -->1}g(u,v,w)abc\times <!--\; \times \; -->dwdudv}$ and I am not sure about g...
To sum up, I appreciate if you could explain me what is going on exactly in a basic way.

Finding the volume in a cylinder that is intersected by a plane
I have a homework question that goes as follows.
Let $V=\{(x,y,z):x^2+y^2\le <!--\; \le \; -->4\text{ and }0\le <!--\; \le \; -->z\le <!--\; \le \; -->4\}$ be a cylinder and let P be the plane through (4, 0, 2), (0, 4, 2), and (-4, -4, 4). Compute the volume of C below the plane P.
So I have these points and I set up my $a(x-<!--\; -\; -->x_0)+b(y-<!--\; -\; -->y_0)+c(z-<!--\; -\; -->z_0)=0$ where $(x_0,y_0,z_0)$ is a point in the plane and ⟨a, b, c⟩ is perpendicular.
I then got two vectors that go between two points, namely:
$PQ=<(0-<!--\; -\; -->4,(4-<!--\; -\; -->0),(2,-<!--\; -\; -->2)>=<-<!--\; -\; -->4,4,0>$ and $PR=<(-<!--\; -\; -->4-<!--\; -\; -->4),(-<!--\; -\; -->4-<!--\; -\; -->0),(4,-<!--\; -\; -->2)>=<-<!--\; -\; -->8,-<!--\; -\; -->4,2>$
I then get the cross product of them to get $<8,8,48>$ which is the coefficient in my equation of the plane:
$8(x-<!--\; -\; -->4)+8(y-<!--\; -\; -->0)+48(z-<!--\; -\; -->2)=0$
I think that this is the equation of my plane. Is it?

Find the DE that describes the change in a raindrops volume over time
A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time
Volume, $V=\frac{4}{3}\mathrm{\pi <!--\; \pi \; -->}{r}^2$ but I dont know how to find any way to relate that to an evaporation rate. This course is an introductory differential equations course and they don't give you anything else to work with.
Do they assume we covered this kind of problem in a previous physics course?
IF the question read:
A spherical raindrop evaporates at a rate proportional to its surface area as $\frac{dV}{dt}=\mathrm{\alpha <!--\; \alpha \; -->}S(t)$. Write a differential equation for the volume of the raindrop as a function of time
I might have fared a bit better.
Is there a way to derive this equation?

Finding volume of sphere cartesian to polar
Using cartesian coordinates the volume of a sphere can be calculated via
${\int <!--\; \int \; -->}_{-<!--\; -\; -->R}^R{\int <!--\; \int \; -->}_{-<!--\; -\; -->\sqrt{R^2-<!--\; -\; -->z^2}}^{\sqrt{R^2-<!--\; -\; -->z^2}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\sqrt{R^2-<!--\; -\; -->z^2-<!--\; -\; -->y^2}}^{\sqrt{R^2-<!--\; -\; -->z^2-<!--\; -\; -->y^2}}1\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}\mathrm{d}\mathrm{z}$
However using a parametrisation with polar coordinates the Integral becomes:
${\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^R{\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}1R^2\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}\mathrm{d}\mathrm{R}\mathrm{d}\mathrm{\phi <!--\; \phi \; -->}$.
Thinking closer about this I just don't see how these integrals are interchangeable. Notably substituting polar coordinates in the cartesian one I don't notice how the borders simplify.
For instance using the substitution
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}R\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->})\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\\ R\sin <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->})\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\\ R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\end{array}\right)$
I get for the first border:
$\sqrt{R^2(1-<!--\; -\; -->{\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->{\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}){\sin }^2<!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}))}$. What's that even?

Volume of Ellipsoid with a hole
How would I go about computing the volume of an ellipsoid with spherical caps removed and a cylindrical hole through it? I'm thinking about finding dV of a cross section (the ellipsoid has a circular horizontal cross section). Would this be the best way to go? I need to find if the volume is dependent on the radius of the sphere.

Constructing a box to find its volume
We want to construct a closed box with a square base. We only have $20m^2$ of material to use in construction of the box. Assuming that all the material is used in the construction process, determine the maximum volume that the box can have to the nearest hundredth of a cubic meter.
Let the dimensions of the base be t while its height is h
Surface area of the box $2t^2+4th=20$
Its volume $t^{2}h$
Substituting value of h into the formula for volume and then finding the derivative to find the value of t and h, gives the answer as 6.09 square meters.
Can someone confirm this answer for me

Need help with the following question about finding volume of improper integral
Find the exact volume of the solid created by rotating the region bounded by $f(x)=\frac{1}{\sqrt{x}\ln <!--\; \; -->x}$ and the x-axis on the interval $[2,\infty )$. State the method of integral used.
My issue specifically is that I have no clue how to integrate the integral because I am using the disk method and the integral ends up being
${\int <!--\; \int \; -->}_2^{\mathrm{\infty <!--\; \infty \; -->}}\mathrm{\pi <!--\; \pi \; -->}\cdot <!--\; \cdot \; -->\frac{1}{x(\ln <!--\; \; -->(x){)}^2}dx$
How would I go about solving this because I am unsure how to integrate that integral.

How to express volume of a sphere as a sum of infinitesimally thick discs?
I want to express the volume of a sphere with a radius r as an integral that adds up each infinitesimally thick disc within the volume. So I have $dV=A(x)dx$, where A(x) is the area of the disc that is at coordinate x. I'm having trouble finding this function in terms of r. Remember, r is the radius of the sphere. At $x=0,A=\pi r^2$, and at $x=r,A=0$. The bounds of the integral should be -r to r I believe. But what is A(x)?

Volume and Triple Integrals
When finding the volume of a solid using triple integrals, is the "integral" always $\int <!--\; \int \; -->\int <!--\; \int \; -->\int <!--\; \int \; -->$ dV, regardless of what the "z function" is?
For instance, for the problem: Find the volume of the solid that lies within the sphere $x^2+y^2+z^2=4$, above the xy-plane, and below the cone $z=$ $\sqrt{x^2+y^2}$.
Is the integral just $\int <!--\; \int \; -->\int <!--\; \int \; -->\int <!--\; \int \; -->$ dV with the correct bounds of integration, regardless of $z=$ $\sqrt{x^2+y^2}$?

Volume of a truncated ellipsoid
My objective is to find the volume of the intersection of the ellipsoid
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le <!--\; \le \; -->1$
and $Ax+By+Cz\le <!--\; \le \; -->D$
where $a,b,c,A,B,C,D\in <!--\; \in \; -->\mathbb{R}$. I have done several attempts but it seems like I have to consider a lot of different cases. My approach is to give a parametrization of the intersection and then finding the volume through solving an integral. However, I can't seem to find a good enough parametrization.

Finding volume of a cone through integration
I am trying to find the volume of a cone using integration through horizontal slicing. The cone has a base radius of 10cm and a height of 5cm.
I am assuming this means I should integrate with respect to y, but I am not entirely sure how to set this up. I know that volume of a cylinder is given by the following:
$V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h$
So I am assuming that the integral would be:
$\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{5}f(y{)}^{2}dy$
I am not sure how the x value of the radius 10cm (since it is not with respect to y) should fit into the equation, though.
Also, sorry for the pseudo-code style. I do not know how to use the math typesetting yet.

Probability of Observing N particles in a given volume?
I'm having an issue with a probability problem concerning solutions.
Assume there is an "observational region" in a dilute solution with a volume V, and as solutes move across its boundary, the number N of solute molecules inside the observation region fluctuates.
Divide V into M regions of volume v each with n particles. The solution is dilute enough that $n=0$ or 1 (there is no v with more than one particle of solute), and each cell is occupied ($n=1$) with probability $p=({\rho }_0)v$.
If W(N) is the number of configurations of the observation volume when N solutes are present, what is the probability P(N) of observing a given value of N, in terms of p,W(N),M, and N.
I know the probability $P(n_1,n_2,\dots <!--\; \dots \; -->,n_M)$ of finding the system in a particular configuration in the observation volume is $p(N)=p^N(1-p{)}^{M-N},$, (Bernoulli Distribution), and since there are N particles in M spaces then the maximum number of configurations is $\frac{M!}{(N!(M-<!--\; -\; -->N)!)}$
I'm not sure where to go from here.

Volume of torus
I wanna calculate the volume of a torus that is created when the circlearea $(x-<!--\; -\; -->4{)}^2+y^2\le <!--\; \le \; -->4$ revolves around the y-axis. I don't understand what the difference would be if I calculated the volume of the circle $(x-<!--\; -\; -->4{)}^2+y^2=4$ revolving around the y-axis. Doesn't the circlearea just create a solid torus instead of a hollow? Then comes the practical problem of finding a formula for the volume. My idea is that if the distance from the "right" and "left" side of the circle to origo is R and r respectively, then the area of the thin circles that constitute the torus is $\mathrm{\pi <!--\; \pi \; -->}(R^2-<!--\; -\; -->r^2)$. Then the height is dy and so we integrate from, if x denotes the radius of the circle, -r to r. Would this work?

Gram Determinant equals volume?
I have been trying to solve this problem of finding the 'n-volume' of a paralleletope spanned by m vectors, where clearly $m\le <!--\; \le \; -->n$. In general, for computational purposes, what I have managed to do is define volume as the product of absolute values of vectors obtained by gram-schmidt orthogonalizationn. (Makes sense right? That's the natural interpretation when we say volume)
I had to do two things, firstly to show that this definition of volume is a well defined one (i.e. any set of orthogonal vectors obtained by the process will give the same volume), and secondly to find a quick way to do this. I managed to prove the first one by induction, but the second part is a little bit of a problem. I managed to obtain formulae for small dimensions as 2,3 or even 4 but this process is impractical for any bigger dimensions as the substitutions for smaller dimensions into the formula for the next dimension becomes exponentially complicated
How does one prove that the gram determinant is equal to the volume of a paralleletope spanned by a set of vectors?

Volume of a Truncated Right Prism with generic base convex polygon
I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}(h_1+h_2+h_3)$, where A is the area of the base and $h_i$ is the height of vertex i of the top face. Or, in other words, to find the volume you "flatten" the top face by finding the average height of its 3 vertices, and then the volume is just that of a regular prism.
My question is: for a generic base convex polygon, does the same argument apply? Is the volume equal to $\frac{A}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{h}_i$, where n is the number of vertices of the base/top? Is there a proof for this anywhere?
My intuition says it's true, but what if the base shape was like a really long kite; wouldn't the height of the vertex at the bottom tip of the kite skew the average height of the top face?

Finding volumes - when to use double integrals and triple integrals?
This is not a technical question at all, but I'm quite confused about what should I use to compute volumes in ${\mathbb{R}}^3$ with integration.
I've read somewhere that a double integral gets the volume swapping across the x and y axis while a triple integral just integrate the whole thing at once, how accurate is this?. Can a volume expressed by a double integral be expressed by a triple integral?, and a triple can always be expressed by a double?

Volume of a parallelepiped, given 8 vertices
Given the eight vertices (0,0,0), (3,0,0), (0,5,1), (3,5,1), (2,0,5), (5,0,5), (2,5,6), and (5,5,6), find the volume of the parallelepiped.
I'm having trouble finding the 1 vertex and 3 vectors needed to find the volume. The closest four vertexes I found so far are (0,0,0),(3,0,0),(0,5,1),(3,5,1)...is using those four vertexes correct? Any starting hints to point me in the right direction?