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Finding the volume of a solid region
I'm trying to find the volume of the solid region inside the sphere $x^2+y^2+z^2=4$, and the upper nappe of the cone $z^2=3x^2+3y^2$ (I only have to set up the triple integral itself, not evaluate it).
So, I need to do this in cylindrical coordinates. Thus, for the r value, since $z^2+y^2=4$, $r=2$. Thus, $0\le <!--\; \le \; -->r\le <!--\; \le \; -->2$.
Next, I need to determine the value of z. I've considered the fact that $y=\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$, and $x=\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$. Thus, $z^2=3({\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))+3({\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))$. This is equivalent to $z^2=3$. Thus, $z=\sqrt{3}$. This would be the lower limit, because it came from the equation of the cone. For the upper boundary of z, I'm not completely sure. Any help with this would be greatly appreciated.
Finally, for $\mathrm{\theta <!--\; \theta \; -->}$, I think it would be 0 to $2\mathrm{\pi <!--\; \pi \; -->}$, since it is the solid region inside a sphere.
I'm not extremely confident in my abilities yet, and I don't want to lead myself astray. Any help/corrections would be greatly appreciated!

Finding volume of solid
Suppose that a solid is formed in such a way that each cross section perpendicular to the x-axis, for $0\le <!--\; \le \; -->x\le <!--\; \le \; -->1$, is a disk, a diameter of which goes from the x-axis out to the curve $y=\sqrt{x}$.
Find the volume of the solid.
For this I use the disk formula. So $\mathrm{\pi <!--\; \pi \; -->}{\displaystyle {\int <!--\; \int \; -->}_0^1(\sqrt{x}{)}^{2}dx.}$.
When I do this, I get ${\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{5}}$. The answer is ${\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{8}}$. What am I doing wrong?

Finding Volume of Rugby Ball
I am asked to find the volume of rugby ball whose surface is given by the ellipsoid:
$\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}=1$
I am having trouble figuring out which coordinate system I should use. Is it possible to solve the triple integral of the volume by just using cartesian co-ordinates, without making conversions to the spherical or cylindrical coordinate system?

Sketching and Finding Volume of a Container
I generally struggle with sketching graphs so some guidance would be appreciated -
A container is made that encloses the volume defined by rotating the curve $x=z^{3/2}$ about the z axis, where $0<z<2$ and x and z are measured in metres. Sketch the cross section of the container in the xz-plane and calculate its volume.
I know I have to somehow find the "main" values of x and z, then sketch from there but i'm confused.

What is the volume in cubic inches of a box that is 25 cm by 25 cm by 25 cm (given 1 inch=2.5 cm approx.)

Finding volume of a sphere using integration
I have searched and found 2 methods of finding volume using integration:
- considering a small cylindrical element and integrating that over the radius
- considering a small circle element and using the relation $x^2+y^2=r^2$ and integrating it over the z-axis.
I was trying to find the integration by considering a small circle element (with radius r) and using the relation $r=R\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ where R is the radius of the sphere / hemisphere.
So I was thinking of calculating the volume of the hemisphere by integrating the $\mathrm{\pi <!--\; \pi \; -->}{R}^2{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}$ from $\mathrm{\theta <!--\; \theta \; -->}$ to $\mathrm{\pi <!--\; \pi \; -->}/2$. Is this method right? And how will the integration be like?

Finding the total surface area of cuboid model using volume
You are to design a cuboid model with a square base that has a volume of $12m^3$. In order to have a minimum total surface area for the cuboid, what are the values for the height of the cuboid and the length of a side in the square base?
So far, this is what I have done:
$\text{Volume = }L.B.H=B^2.H$
$12=B^2.H$
$H=\frac{12}{B^2}$
$\text{S.A. = }2.B^2+4.B.H$
$=2.B^2+4.B.(\frac{12}{B^2})$
$=2.B^2+\frac{48}{B}$

Finding Volume of object in 3D space rotating on x-axis
Find the volume of the solid, generated by revolving the region bounded by $y=\sqrt{\sin <!--\; \; -->(4x)},y=0$ and $0\le <!--\; \le \; -->x\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$ about the x-axis.
To solve for volume, I have the following integral
$V={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}\mathrm{\pi <!--\; \pi \; -->}r(x{)}^{2}dx=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}\sin <!--\; \; -->(4x)dx=\mathrm{\pi <!--\; \pi \; -->}{[-<!--\; -\; -->\frac{\cos <!--\; \; -->(4x)}{4}]}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}.$
That is not one of the solution options. What did I do wrong?

Evaluating the Volume of a Cupola-Shaped Set by Integration
Let $f(x,y)=1-<!--\; -\; -->\frac{x^2}{4}-<!--\; -\; -->y^2$ and $\mathrm{\Omega <!--\; \Omega \; -->}=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:<!--\; :\; -->f(x,y)\ge <!--\; \ge \; -->0\}.$
Compute the volume of the set $A=\{(x,y,z)\in <!--\; \in \; -->{\mathbb{R}}^3:<!--\; :\; -->(x,y)\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->},0\le <!--\; \le \; -->z\le <!--\; \le \; -->f(x,y)\}.$
My idea is to slice the set along the z-axis, obtaining a set $E_z$ - in fact, an ellipse - and computing the volume as ${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_{E_z}dxdydz$.
However, I am stuck finding a way to describe $E_z$. What is the best strategy to do that?

Find the volume of the solid bounded by
$x^2+y^2–2y=0,z=x^2+y^2,z=0$. I have to calculate volume using triple integrals but I struggle with finding intervals. I calculated $x_1=\sqrt{2y-<!--\; -\; -->y^2}$ and $x_2=-<!--\; -\; -->\sqrt{2y-<!--\; -\; -->y^2}$. I think z will have interval from 0 to 2y, but I don't know what to do next.

Finding the volume of an object with 3 parameters
I know how to find the volume of a sphere/ball around x-axis using:
$V=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_a^b{f}^2(x)dx$
Lets say if:
$x^2+y^2=r^2$
We do: $y=\sqrt{r^2-<!--\; -\; -->x^2}$
So now: $V=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_{-<!--\; -\; -->r}^r(\sqrt{r^2-<!--\; -\; -->x^2}{)}^{2}dx$
But the problem starts here:
Im given an equation of ellipsoid, it has 3 parameters:
$x^2+4y^2+4z^2\le <!--\; \le \; -->4$
What do i do with the z parameter? How do i build y now? how does it fit to the equation by integrals of V?
I would like an explanation and not just a solution - because its homework.

A cylinder and a cone have a common base and height. Calculate the volume of the cylinder if the volume of the cone = 10.

The length of a cuboid is 60 cm. Its height is 40% of its length and 3\4 of its width. calculate the volume of the parallelepiped.

Volume of Solid of Revolution about an equation
I learned about the disk and shell method for finding a volume of a solid of revolution. For my question I don't think I can use either method directly. Here is my question:
Find the volume of the region bounded by $y=x$ and $y=x^2$, but rotated about the equation of $y=x$.
Here the axis of revolution is not a vertical or horizontal line, but rather the equation $y=x$. One idea I had was to convert this diagonal line into a horizontal or vertical one, but I can't see to do this. Maybe polar coordinates might be useful, but I'm not sure.

Volume of revolved solid using shell method: finding height
The problem that I am working with is:
Find the volume of the solid of revolution formed by rotating the region R bounded by $y=4+x^2,x=0,y=0,andx=1$ about the line $y=10$.
I have the following so far (using the shell method):
$$\begin{gathered} V={\int <!--\; \int \; -->}_a^{b}2\mathrm{\pi <!--\; \pi \; -->}rhdy \\ r=10-<!--\; -\; -->y \\ c=2\mathrm{\pi <!--\; \pi \; -->}(10-<!--\; -\; -->y) \\ h=? \end{gathered}$$
$V=2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^1(10-<!--\; -\; -->y)hdy$
I've been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.
Would the height just be $\sqrt{y-<!--\; -\; -->4}$ or would it be $10-<!--\; -\; -->\sqrt{y-<!--\; -\; -->4}$ and why?

Susan has 600 feet of fencing to use in making a yard for hernew puppy. The yard she has planned has a length that is 75 feetless than twice the width.
What ae the dimensions of theyard.
The answer that I come up with is Length -75 (W) = yard
150-75 (4) = 4x150
Length=4, Yard= 150
So I don't know if this is right or not.

A backyard garden is shaped like a rectangle with a length that is 25ft longer than the width and a perimeter of 170ft.
What is the width, the length and the area of the garden

Let S be the set of all rectangles with perimeter P. Showthat the square is the element of S with largest area.
I know that $A=x((\frac{P}{2})-<!--\; -\; -->x),\text{ but the next steps is }(\frac{P}{4}{)}^2-<!--\; -\; -->(x-<!--\; -\; -->(\frac{P}{4}){)}^2$, which I dont get.

Find the volume of a solid limited by $64x^2-<!--\; -\; -->4y^2+16z^2=0$ and $y=1$.

Finding the volume using triple integrals
I am curious how I would describe the region $y^2=x$ bounded by the lines $y=1$, $x=1$, $x=0$ and $y=0$ revolved about the y axis in 3 dimensions in order to find the volume in terms of triple integrals. (The solid looks like a cone.) This can easily be done with 1 integral using the "disk" or "shell method, however suppose I don't know those methods and I wanted to find the volume from a very general idea (i.e. triple integrals). What would be the limits of integration in the z plane? Would the limits in the x and y planes simply be from 0 to 1?