Unleash Your Finding Volume Potential with Expert Support and Comprehensive Practice

Let V be volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$.
Express V as a multiple integral, whose limits should be clearly stated. Hence calculate V.
Solution
I want to find out the limits of the multiple integral that's needed to calculate V.
I'm guessing that: $x=a-<!--\; -\; -->z$, $x=(y^2)/4a$.
$y=\pm <!--\; \pm \; -->\sqrt{4ax},z=0,z=a-<!--\; -\; -->x$
But it seems like I've used the upper plane equation twice?
Also, it would really help if we could compare our answer volumes to check that this is right from the start!

Finding a volume of a region defined by $|x-<!--\; -\; -->y+z|+|y-<!--\; -\; -->z+x|+|z-<!--\; -\; -->x+y|=1$.
I'm having trouble approaching this problem. Could someone maybe give me a hint or a solution, it would be so helpful.

Finding the volume of this intersection
How to find the volume of this:
The region common to the interiors of the cylinders $x^2+y^2=1$ and $x^2+z^2=1$ and the first octant.
I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are x, y, z′s in both equations and that I can solve with triple but I dont know what to do here. I started of doing $x^2+z^2=x^2+y^2$ so $z=y$ but dont know what to do with this it doesnt look like the line of intersection is $z=y$ in the diagram given.

Find the volume of the regions enclosed by $z=x^2+y^2-<!--\; -\; -->2$ and $z=30-<!--\; -\; -->x^2-<!--\; -\; -->y^2$
I set up a triple integral with the bounds of the inmost as $x^2+y^2-<!--\; -\; -->2$ to $30-<!--\; -\; -->x^2-<!--\; -\; -->y^2$. The two outer integrals both had the bounds from -4 to 4. When I solved it I got 1024 as the volume, but this isn't correct. Can someone please show me the steps to finding the bounds, and the correct answer?

Volume integral - finding the region
I need to find the volume of an object set with the following function:
$x+y+z=1$
And all three axis.
So I converted the function into $z=-<!--\; -\; -->x-<!--\; -\; -->y+1$, and it gave me kind of a clepsydra, crossing x,y plane in $x=y=1$.
So the region seems to be the circle $x^2+y^2=1$.
However, changing x to r, and y to $\mathrm{\varphi <!--\; \varphi \; -->}$ gives me wrong result $\mathrm{\pi <!--\; \pi \; -->}$, and it should be 1/6.
Is the region I've came up with correct?

Finding value to calculate volume
Given $D=[(x;y)\in <!--\; \in \; -->{\mathbb{R}}^2:1\le <!--\; \le \; -->y\le <!--\; \le \; -->a{x}^2+1,0\le <!--\; \le \; -->x\le <!--\; \le \; -->2/a],0<<!--\; <\; -->a$ let W be the region obtained by rotating D around Y axis.
A) Find the volume of W
B) Find, if possible, the a values $\mathrm{ϵ<!--\; ϵ\; -->}(0;+\mathrm{\infty <!--\; \infty \; -->})$ so that the volume of W is a minimum, and a maximum
C) Find, if possible, the a values $\mathrm{ϵ<!--\; ϵ\; -->}(1/3;3)$ so that the volume of W is a minimum, and a maximum
Well, what I've done so far is finding the inner and outer radius to calculate the volume in terms of Y. The inner radius would be be $\sqrt{\frac{y-<!--\; -\; -->1}{a}}$ and the outer would be $\frac{4+a}{a}$, which is a evaluated in the parabola. Then, the volume would be ${\int <!--\; \int \; -->}_1^{\frac{4+a}{a}}(\sqrt{\frac{y-<!--\; -\; -->1}{a}}{)}^2-<!--\; -\; -->(\frac{4+a}{a}{)}^{2}dx$.
And that's the function that I have to differentiate to find its maxima and minima, which, after differentiating, is 1/a. Is ok what am I doing? How can I go on?

Finding the height and top radius of cone so that volume is maximum and Finding the angle so that the volume is maximum
You have a 6-inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge with central angle theta and forming the remaining paper into a cone. - You are given that 3 is the slant height
a) Find the height and top radius of the cone so that the volume of the cup is as large as possible.
b) What is the angle theta of the arc of the paper that is cut out to create the cone of maximum volume?
I know how to do related rates with volume, but I can't seem to figure it out with angles being cut out from a circle.

Finding the volume of the tetrahedron.
Find the volume of the tetrahedron with the vertices P(1,1,1), Q(1,2,3), R(3,1,2), and S(2,3,1).

Finding Volume Using Cylindrical Shells
It says use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis $y=x^{3}y=8$ about the axis $x=3$. I drew the graph, reflected it about the $x=3$ line, and drew a cylinder. `I figured that the radius is just $r=3-<!--\; -\; -->x$ and the height would just be $h=x^3-<!--\; -\; -->0$ (since the lowest y value is zero), and plugged these into the integral of $2(\mathrm{\pi <!--\; \pi \; -->})(r)(h)$ from 0 to 2. However, I got the wrong answer (correct answer should be $264\mathrm{\pi <!--\; \pi \; -->}/5$).
I have a feeling that my height may be wrong but I'm not sure why.

Finding the volume of a solid of revolution about the x-axis
Let R be the area laying beneath the curve $f(x)=8-<!--\; -\; -->x^2$ and above the line $y=7$. Find the volume of the solid of revolution which is created when R is revolved around the x-axis.
I graphed the functions and found out I have to integrate from -1 to 1. I want to use the disc method, but I don't know how to only get the discs with radii higher than $y=7$ and lower than f(x). I thought it would be logical if the radii of the discs would be $8-<!--\; -\; -->x^2-<!--\; -\; -->7=1-<!--\; -\; -->x^2$, but this approach gives me the wrong answer when i plug it into the formula.

Volume of Solid Enclosed by an Equation
I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:
Compute the volume of the solid enclosed by
1. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,x=0,y=o,z=0$
2. $x^2+y^2-<!--\; -\; -->2ax=0,z=0,x^2+y^2=z^2$

Volume between under parabaloid
Suppose I want to find the volume between $z=2x^2+3y^2$ and $z=4$. Is there a way finding that with a double integral? I tried to use $4-<!--\; -\; -->2x^2-<!--\; -\; -->3y^2$ inside the integral and then convert it to polar coordinates...

Finding volume of solid revolution about x-axis
I'm asked to find the volume of a container of height 13.5 The container is made by rotating $y=1.5x^2$ where $0\le <!--\; \le \; -->x\le <!--\; \le \; -->3$ about the axis $x=-<!--\; -\; -->0.5$ (the bottom is flat).
I tried finding the right volume a number of ways, but I believe my most solid attempt so far is this:
I invert the function to get $y=\sqrt{\frac{2}{3}x}$.
From this I tried to calculate $\mathrm{\pi <!--\; \pi \; -->}\ast <!--\; \ast \; -->{\int <!--\; \int \; -->}_0^3\sqrt{\frac{2}{3}x}dx$ which gave me $3\mathrm{\pi <!--\; \pi \; -->}$ which I thought was correct, but it wasnt. Plotting this, I realized the height becomes 3 when I integrate the inverted function from 0 to 3, so I thought I should maybe integrate from 0 to 13.5. In that case I get $\frac{243\mathrm{\pi <!--\; \pi \; -->}}{4}$ which is also incorrect. I'm sure I've misinterpreted what it means to rotate the function about the axis $x=-<!--\; -\; -->0.5$

Find the volume of a solid which is bounded by the paraboloid $4z=x^2+y^2$, the cone $z^2=x^2+y^2$ and the cylinder $x^2+y^2=2x$
I approached this problem by trying to find the volume bounded by the paraboloid and the cylinder and then subtracting it from the volume bounded by the cone and the cylinder. But I am getting the wrong answer. I converted the all the bounds into cylindrical co-ordinates.
For finding the volume bounded by the cone and the cylinder,
Bounds of integration: $z=0$ to $z=r,r=0$ to $r=2\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}),\mathrm{\theta <!--\; \theta \; -->}=0$ to $\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}$.
For finding the volume bounded by the paraboloid and the cylinder,
Bounds of integration: $z=0$ to $z=\frac{r^2}{4},r=0$ to $r=1,\mathrm{\theta <!--\; \theta \; -->}=0$ to $\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}$.

How can I find the volume of a solid of revolution
Find the volume of the solid obtained by rotating the region bounded by $y=x^2$ and $x=y^2$. Rotating about $y=1$.
I got an intercept of those functions which was (1,1). I tried to use washer method then I got
$\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^1[(x{)}^2-<!--\; -\; -->(\sqrt{x}{)}^2]dx$ and I took integral of the functions but my volume was not right number. I think my way to solve was not right. Could you post correct way to find the volume?

Finding a volume integral in an ellipsoid
I am trying to find the volume integral of $\mathrm{\rho <!--\; \rho \; -->}={\mathrm{\rho <!--\; \rho \; -->}}_0\left(\frac{R^2-<!--\; -\; -->r^2}{R^2}\right)$ inside an ellipsoid given by $\frac{x^2}{(3R{)}^2}+\frac{y^2}{(4R{)}^2}+\frac{z^2}{(5R{)}^2}=1$
I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships $x=3Ru$, $y=4Rv$, $z=5Rw$.
But the resulting integral is still heavy
$\int <!--\; \int \; -->{\mathrm{\rho <!--\; \rho \; -->}}_0(1-<!--\; -\; -->(9u^2+16w^2+25w^2))60dudvdw$
Does anyone have any insight to a more elegant way.

Finding volume of a shape using double integral
I'm trying to find the volume of a given shape:
$V=\{\begin{array}{l}\sqrt{x}+\sqrt{y}+\sqrt{z}\le <!--\; \le \; -->1\\ x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,z\ge <!--\; \ge \; -->0\end{array}$
using double integral. Unfortunately I don't know how to start, namely:
$z=(1-<!--\; -\; -->\sqrt{y}-<!--\; -\; -->\sqrt{x}{)}^2$
and now what should I do? Wolfram can't even plot this function, I'm unable to imagine how it looks like...
Would it be simpler with a triple integral?

Finding the volume of circular cone.
I´m trying to find the Volume of a Circular right cone. The height is 3 units (y-axis) and the radius is 2 units (x-axis).
So, if i Want to find the volume I have to write a equation above in terms of "y", for this situation. Right? The question ask me the equation that generates the solid. I have found $x=(2/3)y$. Am I right?

Revolving a function around y-axis, finding volume using shell method
The region bounded by $y=\frac{x}{\sqrt{x^3+8}}$, the x-axis, and the line $x=2$ is revolved about the y-axis. Find the volume of the solid generated this way.

Finding Volume Using Triple Integration
Set up and evaluate a triple integral to find the volume of the region bounded by the paraboloid $z=1-<!--\; -\; -->\frac{x^2}{9}-<!--\; -\; -->\frac{y^2}{100}$ and the xy-plane.
I understand I'm finding the volume of a paraboloid that forms a "dome" over the xy-plane. Moreover, I can see the paraboloid intersects with the xy-plane to form an ellipse given by $\frac{x^2}{9}-<!--\; -\; -->\frac{y^2}{100}=1$.
I have tried setting this up using rectangular coordinates but the integral started looking extremely messy. I then tried spherical coordinates but had trouble find the upper bound of $\mathrm{\rho <!--\; \rho \; -->}$. Specifically, I can't seem to successfully translate the rectangular equation $z=1-<!--\; -\; -->\frac{x^2}{9}-<!--\; -\; -->\frac{y^2}{100}$ to a spherical equation and isolate $\mathrm{\rho <!--\; \rho \; -->}$.