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Finding the volume of a cube using spherical coordinates
Calculate the volume of a cube having edge length a by integrating in spherical coordinates. Suppose that the cube have all the edges on the positive semi-axis. Let us divide it by the plane passing through the points (0;0;0),(0;0;a),(a;a;0).
We get two equivalent prism; now we divide the section again by the plane passing through the points (a;0;a),(0;0;0),(a;a;a); So one may write:
$\frac{1}{2}V=({\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}d\mathrm{\varphi <!--\; \varphi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}d\mathrm{\theta <!--\; \theta \; -->}{\int <!--\; \int \; -->}_0^{\frac{a}{\cos <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}}}{r}^2\sin <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}dr+{\int <!--\; \int \; -->}_{\mathrm{\pi <!--\; \pi \; -->}/4}^{\mathrm{\pi <!--\; \pi \; -->}/2}d\mathrm{\varphi <!--\; \varphi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}d\mathrm{\theta <!--\; \theta \; -->}{\int <!--\; \int \; -->}_0^{\frac{a}{\sin <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}{r}^2\sin <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}dr)$
So one should expect $V=a^3$ but the latter expression gives a different result. What is the correct way to calculate V and why doesn't my reasoning work?

About finding the rotational volume
A closed area of y-axis, $y=\mathrm{\pi <!--\; \pi \; -->},y=x+sin(x)$ is rotated on line $y=x$. Find the volume of it.
So, I think I have to do an integral looks like this.
$V={\int <!--\; \int \; -->}_0^{\sqrt{2}\mathrm{\pi <!--\; \pi \; -->}}\mathrm{\pi <!--\; \pi \; -->}{r}^{2}dt$, where r is the radius of disk and t is a variable on line $y=x$.
How should I set up the integral and change the variables?

Finding the volume of solid
Use triple integral to find the volume of the solid enclosed between the elliptic cylinder $x^2+9y^2=9$ and the planes $z=0$ and $z=x+3$

Finding volume of a box
You are required to construct an open box from a paper by cutting four squared from the corners of the paper. The rectangular piece of paper is 5cm long and 3cm wide. What should be the length of the sides of such four (identical) squares that will create a box of largest volume? What will be the largest volume?

Finding volume of region with cross sections
The base of a solid is the region is between the line of $y=4$ and the parabola $y=x^2$ the cross sections of the solid that are perpendicular to the x-axis are semicircles. What is the volume?
So the question I have on this is whether or not I'm going to need to divide the integral by 2 since they are semi circles, not circles. When I divide by 2, I get $\frac{16\pi }{3}$; however, when I don't divide, I get $\frac{512\pi }{15}$. Are either of these correct, and if so, which one?

So, the volume of the solid of revolution is
Volume = $V=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_a^b(9-<!--\; -\; -->x^2{)}^{2}dx.$
To find the volume, integrate from x =??? to x = 3

Finding volume of the solid
The region bounded by the curve $y=\sqrt{x}$, the x-axis and the line $x=4$ is revolved about y-axis to generate a solid. Find the volume of solid.
I found $32\mathrm{\pi <!--\; \pi \; -->}/5$? But answer is wrong. Where did I wrong?
${\text{Volume}}_y=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^2[A(y){]}^{2}dy=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^2{y}^{4}dy$

Integrating a specific integral based on the volume of the intersection of two cylinders
I have been able to obtain the formula to calculate the volume like the solution to the link above.
My formula looked like this:
$V={\int <!--\; \int \; -->}_{-<!--\; -\; -->r}^r\sqrt{r^2-<!--\; -\; -->y^2}\sqrt{R^2-<!--\; -\; -->y^2}dy$
Where $r\le <!--\; \le \; -->R$
if $b=\frac{R}{r}$, how can I show $V=r^{3}F(b)$
I am having trouble finding an expression where volume is equal to $r^3$ times some expression that is only dependant on b

Finding volume by rotation of a triangle
The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. Problem:Find the volume generated by revolving the following region: The triangle with vertices (1,1), (1,2) and (2,2) about the y-axis.
Answer:
Since we are going around the y-axis, our integral will integrate with respect to y not x. Observe that the points (1,1) and (2,2) are on the line $y=x$. Let v be the volume we seek.
$\begin{array}{rl}v& =\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_1^2{2}^2-<!--\; -\; -->y^{2}dy=\mathrm{\pi <!--\; \pi \; -->}(4y-<!--\; -\; -->\frac{y^3}{3}){|}_1^2\\ v& =\mathrm{\pi <!--\; \pi \; -->}(8-<!--\; -\; -->\frac{8}{3}-<!--\; -\; -->(4-<!--\; -\; -->\frac{1}{3}))=\mathrm{\pi <!--\; \pi \; -->}(8-<!--\; -\; -->\frac{8}{3}-<!--\; -\; -->4+\frac{1}{3})\\ v& =\mathrm{\pi <!--\; \pi \; -->}(4-<!--\; -\; -->\frac{7}{3})\\ v& =\frac{5\mathrm{\pi <!--\; \pi \; -->}}{3}\end{array}$
The book's answer is $\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3}$. What did I do wrong?

Finding the volume of a sphere with a triple integral and trig sub
How is trigonometric substitution done with a triple integral? For instance, $8{\int <!--\; \int \; -->}_0^r{\int <!--\; \int \; -->}_0^{\sqrt{r^2-<!--\; -\; -->x^2}}{\int <!--\; \int \; -->}_0^{\sqrt{r^2-<!--\; -\; -->x^2-<!--\; -\; -->y^2}}(1)dzdydx$
Here the limits have been chosen to slice an 8th of a sphere through the origin of radius r, and to multiply this volume by 8. Without converting coordinates, how might a trig substitution be done to solve this?

Finding volume of convex polyhedron given vertices
I am trying to compute the volume of the convex polyhedron with vertices (0,0,0), (1,0,0), (0,2,0), (0,0,3), and (10,10,10). I am supposed to use a triple integral but am struggling with how to set it up.

Finding the volume bounded by two equations rotated around $y=2$
I need to find the volume of a solid generated by a rotating the area bounded by the equations about $y=2$:
- $y=1$
- $y=x^2$
I've graphed the functions, but I'm not sure how to setup an integral to find the volume.

Finding radius for volume of revolution
I'm getting comfortable immediately recognizing the outer and inner radius when the axis of rotation is adjacent to the function or when my visible functions are strictly curves.
Take for example the function $f(x)=8x^3$ bounded by $y=0,x=1$ rotated about the line $x=2$.
Geometrically I think the outer radius R is the x distance between f(x) and $x=2$. Likewise, the inner radius r is the x distance between the lines $x=1$ and $x=2$. However, with there being a gap and with constants involved, I'm having difficulty understanding how to determine inner and outer radius. What I did was say that
$R=f(y)+1,r=1$
However, according to Symbolab, $R=f(y)-<!--\; -\; -->2$ and $r=1-<!--\; -\; -->2$.
I don't understand how this works because it appears there is a negative radius, and then it appears the big radius doesn't go up to center of the hole of the solid. I thought the radius of a washer is from edge to center rather than outer edge to inner edge.
I understand the purpose of everything else. I know volume is only interested in the actual object and not the empty space caused by the gap in our function, so I understand why we're subtracting the area of the little circle. I understand why we're summing up volume slice by slice via integration. I understand that dy means we're summing vertically. Etc. The one thing I don't understand is how radius is obtained the way it is.

Volume generated by lemniscate revolving about a tangent at the pole.
The lemniscates $r^2=a^2\cos <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}$ revolves about a tangent at the pole. What is the volume generated by it ?
Please explain in detail. I found a couple of answers on finding surface areas, volumes but didn't understood it.
Can someone solve this. I'm stuck in the middle with integrations.

Finding volume of a frustum of a pyramid
I need to find the volume of a frustum of a pyramid with square base of side b, square top of side a, and height h(using integrals). I have no idea how to do questions like these, I only know to use the disc/washer/cylindrical shells methods and rotate a region around any line and find its volume. For these type of question, I find myself at a loss as to where to even start. Any hints on general about starting these kind of problems are also appreciated!

Finding the volume of a rectangular-based pyramid using calculus?
Find the volume of a pyramid with height h and rectangular base with dimensions b and 2b.

Calculating limits of integrals of volume
I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
$h(x,y)=x{y}^2,0\le <!--\; \le \; -->x\le <!--\; \le \; -->1,1\le <!--\; \le \; -->y\le <!--\; \le \; -->2$
$\therefore <!--\; \therefore \; -->V={\int <!--\; \int \; -->}_1^2{\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{x{y}^2}dzdxdy={\int <!--\; \int \; -->}_1^2{\int <!--\; \int \; -->}_0^{1}x{y}^{2}dxdy=\frac{1}{2}{\int <!--\; \int \; -->}_1^2{y}^{2}dy=\frac{7}{6}$
Step 2
However, on later questions the boundaries are given as:
$h(x,y)=e^{-<!--\; -\; -->x-<!--\; -\; -->y},0\le <!--\; \le \; -->x,y<\mathrm{\infty <!--\; \infty \; -->}$
which to me implies: $V={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_0^{\exp <!--\; \; -->(-<!--\; -\; -->x-<!--\; -\; -->y)}dzdxdy.$
However this evaluates to: $V={\left[e^y\right]}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}.$ which is clearly divergent.
How should I calculate what the limits are for this volume?

Basic questions about finding a volume formula for a box inside of an ellipsoid,
I want to maximize the volume of a box, with sides parallel to the xy, xz and yz-planes, with the box inside of an ellipsoid
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$
So, my answer, using Lagrange multipliers, turns out to be wrong, at least comparing my work to the solution given.
I had thought to maximize "length times height times width", so I figured that the objective function should be $f(x,y,z)=xyxzyz=x^2{y}^2{z}^2$.
The solution instead maximizes a different function:
$f(x,y,z)=2x2y2z$
I am guessing that the solution is indeed correct? And the point I probably missed was this: the ellipsoid is centered at 0. So, sketching out the box on paper, "length times height times width" does look like $2x2y2z$.
What do you think?
Also, how does an equation of an ellipsoid not centered at 0 look like? Would it be something like this:
$\frac{(x-<!--\; -\; -->1{)}^2}{a^2}+\frac{(y-<!--\; -\; -->2{)}^2}{b^2}+\frac{(z-<!--\; -\; -->3{)}^2}{c^2}=1?$
would this be an ellipsoid centered at (1,2,3)?

Help with finding the volume of a solid of revolution
Revolve $x^2+4y^2=4$.
a. About $y=2$.
b. About $x=2$.
The answer at the back of the book is the same for both
$a.2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{2}8\sqrt{1-<!--\; -\; -->\frac{x^2}{4}}dx=78.95684$
$b.2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{1}8\sqrt{4-<!--\; -\; -->4y^2}dy=78.95684$
I tried splitting the ellipse into its upper and lower half but I still couldn't get the answer. I get that its symmetrical that's why it's multiplied to two but I don't know how you get the equation inside the integral.

Finding the volume of liquid in an inclined cylinder
A right circular cylinder is at an incline of ${15}^{\circ <!--\; \circ \; -->}$ from the horizontal and the liquid is level with the lowest point of the top rim of the can. The radius is 3.2004 cm and the height is 11.938 cm. What is the volume of the liquid?
I believe I should use integration with cross-sections of rectangles. The width of each rectangle would be the diameter of the cylinder. I believe my limits of integration would be from 0 to 4.7 \cdot \sin(15^{\circ}) or 1.2164. I'm not sure how to figure out the changing lengths of the rectangles.
Am I on the right track?
${\int <!--\; \int \; -->}_0^{4.7\sin <!--\; \; -->(15°)}6.4008?dy$