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A cylindrical container has a base area 100m square and is 12m high if the container is one-third thid filled with water, what is volume of the water in the container? wich one is it is it A300m or B400m or C600 mor D1200m wich one is correct.

Calculate the volume of a regular pyramid of height h
Calculate the volume of a regular pyramid of height h, knowing that this pyramid is based on a convex polygon whose sum of inner angles is nπ and the ratio between the lateral surface and the base area is k.
Idea: Use the formula $\boxed{{\displaystyle \frac{lwh}{3}}}$ for finding the volume. But at first you have to Bash the problem to find the width and length and then after getting their values you have to plug it into the formula to find the volume.

Finding volume via integration
When we are asked to find the volume for example of a sphere, we slice it into numerous disks with small depth dr and get the infinitesimal volume dV. And then we sum those slices together by putting an integral symbol. But my question is the basic integration we learnt was to find area under the curve.In that case we proved that the area is simply the anti derivative of the function. But here how do we use integration by putting an integral symbol before Adr without proving that the derivative of the volume is the area of the disks? I am so disturbed by this as I didn't see any author giving the proof with proper intuition.

How to calculate the volume of cos(x) around the x-axis and the y-axis separately?
I have difficulties finding the right formula how to calculate the volume rotating cos(x) around the x-axis and y-axis separately can you please give me a hint how to do it?
The interval is $x=[-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$

Finding volume of a solid about a vertical line
How can I calculate the volume of the solid generated by rotating the region bounded by
$y=x^2-<!--\; -\; -->6x+10$
and $y=-<!--\; -\; -->x^2+4x+2$.
About the line $x=4$

Finding volume of a sphere
I am stuck on the following problem:
A circle $x^2+y^2=a^2$ is rotated around the y-axis to form a solid sphere of radius a. A plane perpendicular to the y-axis at $y=\frac{a}{2}$ cuts off a spherical cap from the sphere. What fraction of the total volume of the sphere is contained in the cap?
So far I have figured out the following:
Rotating the cap on the y axis we get a height h starting from $y=\frac{a}{2}$. The interval from $y=0$ to $y=\frac{a}{2}$ (the region below the cap) should be:
$a-<!--\; -\; -->h$
I also know that the radius of the sliced disk, x, can be derived from the equation of the circle:
$x=\sqrt{a^2-<!--\; -\; -->y^2}$
Since the area of a circle is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$ the area with respect to y for the circle should be:
$A(y)=\mathrm{\pi <!--\; \pi \; -->}(a^2-<!--\; -\; -->y^2)$
So to find the volume, we need to integrate the function:
$V={\int <!--\; \int \; -->}_{\frac{a}{2}}^a\mathrm{\pi <!--\; \pi \; -->}(a^2-<!--\; -\; -->y^2)dy$
I know where I should go, but I am not sure what to do about the constraint $y=\frac{a}{2}$ at this point. Should I integrate the terms with respect to y first and then plug in the value which is equal to y? Or should this be done before integrating?

Problem with finding the volume with cylindrical coordinates.
I have the following function:
$Z=x^2+y^2,Z=x+y$
that i want to solve (to find the volume) with cylindrical coordinates. I am evaluating the integral to get:
$V=\iiint <!--\; \iiint \; -->dV=\underset{R}{\iint <!--\; \iint \; -->}{\int <!--\; \int \; -->}_{x^2+y^2}^{x+y}dzdA=\underset{R}{\iint <!--\; \iint \; -->}x+y-<!--\; -\; -->(x^2+y^2)dA.$
and from here i am trying to get the bounds for r by intersecting the two functions and i get that $r=0orr=1$ therefore i tought that $0\le <!--\; \le \; -->r\le <!--\; \le \; -->1$ but it dosen't seems right. But i get the following result:
$2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^1(r-<!--\; -\; -->r^2)rdrd\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}(\frac{r^3}{3}-<!--\; -\; -->\frac{r^4}{4}){|}_0^1=\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$
and this is not the corrent answer, i should get $\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}.$. I know this is too specific to a given problem question i apologize for that but can someone tells me where i made a mistake

Finding the volume bounded by surface in spherical coordinates
$R=4-<!--\; -\; -->1\cos <!--\; \; -->(\mathrm{\varphi <!--\; \varphi \; -->})$

Using Differentials to Calculate the Volume of a Square Pyramid
Use differentials to solve the problem:
The Louvre Pyramid is a tourist attraction in Europe. It is a square pyramid, with a height of 21m, and base of side length 35m. The four faces of this pyramid are covered in glass, of thickness 0.03m. Find the volume of glass used to construct the exterior of the Louvre.
I know that the volume of a square pyramid is: $V=\frac{a^{2}h}{3}$, where a is its base length and h is its height.
I then solved for $dV=\frac{a^{2}dh+2ahda}{3}$, but I am stuck in this equation because there are many unknowns.
The next step I can think of is finding the volume, which can be in the form of: $V_{Glass}=V(Value+0.03)-<!--\; -\; -->V(Value)$, which can be evaluated like: $L(x)=f(x_0)+f^{\prime }(x_0)(x-<!--\; -\; -->x_0)$, but I am not sure of the way of finding the V(Value).

Finding volume of solid using shell method.
Rotating the region bounded by $y=x^3,y=0,x=2$ around the line $y=8$.
I just want to double check if my initial formula is on the right track.
$V_y={\int <!--\; \int \; -->}_0^8(8-<!--\; -\; -->y)(y^{\frac{1}{3}})dy$

Finding the volume of a region rotated about the y-axis.
I'm having trouble trying to find the volume of the region formed by $y=x^2-<!--\; -\; -->7x+10$ and $y=x+3$ rotated about the y-axis. I was able to graph it, but I'm having difficulty when trying to come up with the integral for it.
Any help showing me how to get the integral I should use would be a huge help.

Finding the maximum possible volume of a tetrahedron
Suppose I have three vectors $\stackrel{¯<!--\; ¯\; -->}{x}=a\stackrel{^<!--\; ^\; -->}{i}+b\stackrel{^<!--\; ^\; -->}{j}+c\stackrel{^<!--\; ^\; -->}{k}$, $\stackrel{¯<!--\; ¯\; -->}{y}=b\stackrel{^<!--\; ^\; -->}{i}+c\stackrel{^<!--\; ^\; -->}{j}+a\stackrel{^<!--\; ^\; -->}{k}$, $\stackrel{¯<!--\; ¯\; -->}{z}=c\stackrel{^<!--\; ^\; -->}{i}+a\stackrel{^<!--\; ^\; -->}{j}+b\stackrel{^<!--\; ^\; -->}{k}$, where we can choose a, b, c from the set {1,2,3,....13}. We have to find the probability of the tetrahedron formed by $\stackrel{¯<!--\; ¯\; -->}{x}$, $\stackrel{¯<!--\; ¯\; -->}{y}$, $\stackrel{¯<!--\; ¯\; -->}{z}$ having maximum volume. How should I maximise the determinant for finding the maximum possible volume of the resulting tetrahedron? Can someone help me figure this out?

Finding the volume of $f(x,y,z)=z$ inside the cylinder and outside the hyperboloid
I need to setup the triple integral in cartesian coordinates to solve the volume of $f(x,y,z)=z$ inside the cylinder $x^2+y^2=10$ and outside the hyperboloid $x^2+y^2-<!--\; -\; -->z^2=1$ in the first octant.
What I did so far is to find the intersection between the cylinder and hyperboloid so that I can find the bounds for z. Actually, I am uncertain if finding this intersection is a right procedure.
$10-<!--\; -\; -->z^2=1⟹<!--\; ⟹\; -->z=3$
My bounds for z is $0\le <!--\; \le \; -->z\le <!--\; \le \; -->3$ since the volume is also bounded by the first octant. My bounds for y is $\sqrt{1-<!--\; -\; -->x^2}\le <!--\; \le \; -->y\le <!--\; \le \; -->\sqrt{10-<!--\; -\; -->x^2}$. Lastly, my bounds for x is $1\le <!--\; \le \; -->x\le <!--\; \le \; -->\sqrt{10}$. My iterated integral for this one will be like this ${\int <!--\; \int \; -->}_1^{\sqrt{10}}{\int <!--\; \int \; -->}_{\sqrt{1-<!--\; -\; -->x^2}}^{\sqrt{10-<!--\; -\; -->x^2}}{\int <!--\; \int \; -->}_0^{3}zdzdydx$
Do I miss something here?

Finding value of t such that volume contained inside the planes is minimum
The question is to find out the value of t such that volume contained inside the planes $\sqrt{1-<!--\; -\; -->t^2}x+tz=2\sqrt{1-<!--\; -\; -->t^2},$
$z=0,$
$x=2+\frac{t\sqrt{4t^2-<!--\; -\; -->5t+2}}{\sqrt{12}(1-<!--\; -\; -->t^2{)}^{1/4}}\text{ and }$
$|y|=2$ is maximum.
I tried to figure out the line of the intersection of the planes but it made the problem more complicated.Is there any way by which the equations can be decomposed to make it easier to handle?Any help shall be highly appreciated.

Finding the Volume of a solid - Application of Integrals - Exercise that is not clear to understand
I've been working on a few exercises and one of them seems not clear, I'm not sure what the author meant in it. Here's the exercise:
Find the volume of the solid whose base is the area between the curve $\begin{array}{rl}y& =x^3\end{array}$ and the y axis, from $x=0$ to $y=1$, considering that his cross sections, taken perpendicular to the y axis, are squares.

How to find the volume by using shell method
I would like to find the volume of $\frac{x^2}{9}+y^2=1$ rotated around the x axis, but using the shell method. How can I do this?
Letting the thickness be dy and the problem I encountered is finding the radius.

Finding the volume under the plane $x+y+z=1$
My two methods are giving me different answers, and I'm not sure why.
I am more confident in: ${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x}{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x-<!--\; -\; -->y}dzdydx=\frac{1}{6}$.
But another way I'm trying is to sum the area of all the equilateral triangles underneath, and parallel to, the plane. The side lengths of these triangles range from 0 to $\sqrt{2}$.
Thus: ${\int <!--\; \int \; -->}_0^{\sqrt{2}}{s}^2\frac{\sqrt{3}}{4}ds=\frac{1}{\sqrt{6}}$
I am pretty sure the second approach is incorrect, but why?
We are given $x,y,z>0$

Volume using shells
I'm working on a problem of finding volume between two functions using the shell method. The functions given are $f(x)=2x-<!--\; -\; -->x²$ and $g(x)=x$. It is reflected across the x-axis.
I solved this previously using washers but the problem asks to solve using two methods. I believe this is a dy problem, thus I am trying to convert the two functions into f(y) and g(y), but I don't understand how to convert f(x) into an f(y). It doesn't seem like it can be solved in terms of y. Maybe I am missing something after looking at this too long.
How would I solve this problem using shells? I am approaching this correctly?

How do I know which method of revolution to use for finding volume in Calculus?
Is there any easy way to decide which method I should use to find the volume of a revolution in Calculus? I'm currently in the middle of my second attempt at Calculus II, and I am getting tripped up once again by this concept, which seems like it should be rather straight forward, but I can't figure it out. If I have the option of the disk/washer method and the shell method, how do I know which one I should use?

Finding volume of enclosed region
The base of S is the region enclosed by the parabola $y=9-9x^2$ and the X - axis. Cross-sections perpendicular to the X - axis are isosceles triangles with height equal to the base.