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Finding a rate of sphere area increase given its volume increase
Volume of sphere, $V={\displaystyle \frac{4}{3}}\mathrm{\pi <!--\; \pi \; -->}{r}^3$
Surface area of sphere $S=4\mathrm{\pi <!--\; \pi \; -->}{r}^2$.
If we know, ${\displaystyle \frac{dV}{dt}}=R$.
Let us consider both volume and area as composite functions, thus ${\displaystyle \frac{dV}{dt}}={\displaystyle \frac{dV}{dr}}\times <!--\; \times \; -->{\displaystyle \frac{dr}{dt}}=4\mathrm{\pi <!--\; \pi \; -->}{r}^2\times <!--\; \times \; -->{\displaystyle \frac{dr}{dt}}=R$
whence ${\displaystyle \frac{dr}{dt}}={\displaystyle \frac{R}{4\mathrm{\pi <!--\; \pi \; -->}{r}^2}}$ since ${\displaystyle \frac{dS}{dt}}=6\mathrm{\pi <!--\; \pi \; -->}r{\displaystyle \frac{dr}{dt}}$, let the value of the ${\displaystyle \frac{dr}{dt}}$ into the second equation, to get the answer. Is this approach logically correct?

Finding the maximum volume of a cylinder.
The cylinder is open topped and is made from $200{\text{cm}}^2$ of material.
So what I have done is said $S=2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h+2\mathrm{\pi <!--\; \pi \; -->}{r}^2$.
$\begin{array}{rl}2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h+2\mathrm{\pi <!--\; \pi \; -->}{r}^2& =200\\ 2\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h& =200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2\\ h& =\frac{200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2}{2\mathrm{\pi <!--\; \pi \; -->}{r}^2}\end{array}$
Subbing this into the volume equation I get $V=(\mathrm{\pi <!--\; \pi \; -->}{r}^2)\frac{200-<!--\; -\; -->2\mathrm{\pi <!--\; \pi \; -->}{r}^2}{2\mathrm{\pi <!--\; \pi \; -->}{r}^2}$.
Having trouble differentiating that equation.

Finding dimensions of a rectangular box to optimize volume
A rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of $16f{t}^2$. Compute the dimensions of the box that will maximize its volume.
I am going wrong somewhere but I can't see where. I have set $2(3h+3w+hw)=16$ and my optimization equation is $v=hwl$. I set $w=\frac{8-<!--\; -\; -->3h}{3+h}$ and plug into the optimization equation. Calculating v' gives me $\frac{9h^2-<!--\; -\; -->84h+72}{{(3+h)}^2}$,but setting that equal to zero does not give me the answer. I need for h, which is 1.

Computing volume using shell resulted in negative value?
The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-<!--\; -\; -->x^2$, revolved about $x=2$.
$x^2=2-<!--\; -\; -->x^2\to 2x^2=2\to \frac{2x^2}{2}=\frac{2}{2}\to x^2=1$
$x=1,x=-<!--\; -\; -->1$
Radius is $r=2-<!--\; -\; -->x$
Height is $x^2-<!--\; -\; -->(2-<!--\; -\; -->x^2)$
Finding the Integral, ${\int <!--\; \int \; -->}_{\mathrm{\neg <!--\; \neg \; -->}1}^{1}2\pi (2-<!--\; -\; -->x)(x^2-<!--\; -\; -->(2-<!--\; -\; -->x^2))dx=2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})+C$
Finding the limits, $\underset{x\to -<!--\; -\; -->1+}{lim}2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})=2\pi (\frac{4(-<!--\; -\; -->1{)}^3}{3}-<!--\; -\; -->4(-<!--\; -\; -->1)-<!--\; -\; -->\frac{2(-<!--\; -\; -->1{)}^4}{3}+2(-<!--\; -\; -->1{)}^2-<!--\; -\; -->(-<!--\; -\; -->1{)}^2+\frac{(-<!--\; -\; -->1{)}^4}{6})=19.897$
$\underset{x\to 1-<!--\; -\; -->}{lim}2\pi (\frac{4x^3}{3}-<!--\; -\; -->4x-<!--\; -\; -->\frac{2x^4}{3}+2x^2-<!--\; -\; -->x^2+\frac{x^4}{6})=2\pi (\frac{4(1{)}^3}{3}-<!--\; -\; -->4(1)-<!--\; -\; -->\frac{2(1{)}^4}{3}+2(1{)}^2-<!--\; -\; -->(1{)}^2+\frac{(1{)}^4}{6})=-<!--\; -\; -->13.614$
And finally, computing the volume from the obtained limits $V=-<!--\; -\; -->13.614-<!--\; -\; -->19.897=-<!--\; -\; -->33.510$.
So, if someone could let me know if (or where) I went wrong, that would be great.

Volume of a Solid of Revolution?
Find the volume if the region enclosing $y=x^3,x=0,$, and $y=8$ is rotated about the given line. So the axis of rotation that was given was the y axis, but I'm a bit confused on what the given bounds of $y=8$ and $x=0$ have to do with it, but I pretty much ignored them and put the given equation in terms of y and used the integral ${\int <!--\; \int \; -->}_0^7(2^2-<!--\; -\; -->(\sqrt[3]{y}{)}^2)$ to solve, is this at all right? Any help would be appreciated Also looking for some guidance on how I'd go about finding the volume if the axis of rotation was something like $x=6$.

Conceptual doubt in taking elements for integration
In physics, we were taught to find the center of mass using calculus for different shapes. I tried to find the volume of a sphere using similar concepts.
I took a cylindrical element to integrate whose radius is $R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ where $\mathrm{\theta <!--\; \theta \; -->}$ is the angle made with the vertical to the edge of the element . And took the height to be $Rd\mathrm{\theta <!--\; \theta \; -->}$ when I integrated all the cylindrical elements varying theta from o, $\mathrm{\pi <!--\; \pi \; -->}$ and did not get the right answer.
I realize the error is taking $Rd\mathrm{\theta <!--\; \theta \; -->}$ as height but why is it that we can take the "extension" to be $rd\mathrm{\theta <!--\; \theta \; -->}$ when finding the surface area of the sphere.
To state my question more clearly:
Why cant the height of the element be $Rd\mathrm{\theta <!--\; \theta \; -->}$ when finding volume since it works when finding surface area.
What I did to find volume:
${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\mathrm{\pi <!--\; \pi \; -->}(R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2(Rd\mathrm{\theta <!--\; \theta \; -->})$
What I did to find surface area:
${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}2\mathrm{\pi <!--\; \pi \; -->}R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}(Rd\mathrm{\theta <!--\; \theta \; -->})$

Finding the volume using double integrals
Find the volume of the wedge sliced from the cylinder $x^2+y^2=1$ by the planes $z=a(2-<!--\; -\; -->x)$ and $z=a(x-<!--\; -\; -->2)$.
I am confused because $x^2+y^2=1$ is a unit circle not a cylinder. and the other two are lines in the zx plane where $y=0$. I don't see how they are planes, are they planes or are they just lines in the zx plane (when $y=0$).

Triple integrals-finding the volume of cylinder.
Find the volume of cylinder with base as the disk of unit radius in the xy plane centered at (1, 1, 0) and the top being the surface $z=((x-<!--\; -\; -->1{)}^2+(y-<!--\; -\; -->1{)}^2{)}^{3/2}.$.
I just knew that this problem uses triple integral concept but dont know how to start. I just need someone to suggest an idea to start. I will proceed then.

Volume Integration of Bounded Region
I'm trying to integrate this volume in spherical and cylindrical coordinates, but having difficulty finding my bounds of integration;
I'm given the region in the first octant bounded by $z=$ $\sqrt{x^2+y^2}$, $z=$ $\sqrt{1-<!--\; -\; -->x^2-<!--\; -\; -->y^2}$, $y=x$ and $y=$ $\sqrt{3}x$ and I need to evaluate ${\iiint <!--\; \iiint \; -->}_{V}dV$.
When proceeding to integrate with spherical and cylindrical coordinates I am not getting the right bounds such that both methods equate to the same volume? I am definitely missing something. Any and all advice would be much appreciated!

Finding the volume of oil in a cylinder which is lying parallel to the ground.
A cylinder of height 2 m and radius 2 m is partially filled with oil. When the cylinder is lying parallel to the ground the height of oil level is 1.5 m from ground. What is the volume of oil? How to solve it?

Finding the volume of the tetrahedron with vertices (0,0,0), (2,0,0), (0,2,0), (0,0,2). I get 8; answer is 4/3.
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors A, B and C then the area of the object is $A\cdot <!--\; \cdot \; -->(B\times <!--\; \times \; -->C)$. Let V be the volume we are trying to find.
$\begin{array}{rl}{x}^2& =6-<!--\; -\; -->y^2-<!--\; -\; -->z^2\\ A& =(2,0,0)-<!--\; -\; -->(0,0,0)=(2,0,0)\\ B& =(0,2,0)-<!--\; -\; -->(0,0,0)=(0,2,0)\\ C& =(0,0,2)-<!--\; -\; -->(0,0,0)=(0,0,2)\\ V& =\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\left|\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right|\\ & =2\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=2(4-<!--\; -\; -->0)\\ & =8\end{array}$
However, the book gets $\frac{4}{3}$

Volume of Generalized Tetrahedron in $R^n$.
I'm having difficulty finding the volume of a tetrahedron in ${\mathbb{R}}^n$.
Find the volume of a generalized tetrahedron in ${\mathbb{R}}^n$ bounded by the coordinate hyperplanes and the hyperplane $x_1+x_2+...+x_n=1$
In two dimensions, we have ${\int <!--\; \int \; -->}_0^{1}1-<!--\; -\; -->x_{1}d{x}_1$. In three dimensions, I got something like ${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x_1}1-<!--\; -\; -->x_{2}d{x}_{2}d{x}_1$.
I am off to a good start?

Finding the volume of the tetrahedron using triple integrals
D is the tetrahedron bounded by the coordinate planes and the plane $3x+3y+z=3$, then express the volume of D as a triple integral.
My Try:
The z-limits are $0\le <!--\; \le \; -->z\le <!--\; \le \; -->3-<!--\; -\; -->3x-<!--\; -\; -->3y$
If y is the limit on the xy- plane then $z=0$ $0\le <!--\; \le \; -->y\le <!--\; \le \; -->1-<!--\; -\; -->x$.
Similarly, $0\le <!--\; \le \; -->x\le <!--\; \le \; -->3$.
Finally, the integral will be ${\int <!--\; \int \; -->}_0^3{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x}{\int <!--\; \int \; -->}_0^{3-<!--\; -\; -->3x-<!--\; -\; -->3y}dzdydx$

Finding the volume by Shell method: $y=x^2,y=2-<!--\; -\; -->x^2$, $y=x^2,y=2-<!--\; -\; -->x^2$ about the line $x=1$.
what I get from this after graphing is:
$2\mathrm{\pi <!--\; \pi \; -->}\int <!--\; \int \; -->(1-<!--\; -\; -->x)(2-<!--\; -\; -->2x^2)dx$
which becomes: $2\mathrm{\pi <!--\; \pi \; -->}\int <!--\; \int \; -->(2-<!--\; -\; -->2x^2-<!--\; -\; -->2x+2x^3)dx$
integrating that I get:
$2\mathrm{\pi <!--\; \pi \; -->}[2x-<!--\; -\; -->\frac{2}{3}{x}^3-<!--\; -\; -->x^2+\frac{1}{2}{x}^4]$ from 0 to 1
My answer is $\frac{5}{3}\mathrm{\pi <!--\; \pi \; -->}$ but my book says the answer is: $\frac{16}{3}\mathrm{\pi <!--\; \pi \; -->}$
Could someone tell me where I went wrong? Was it the upper lower bounds?

AP Calc AB Problem - Finding volume
The base of a solid in the region bounded by the two parabolas $y^2=8x$ and $x^2=8y$. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?
The answer choices are:
A) $\frac{288\mathrm{\pi <!--\; \pi \; -->}}{35}$
B) $\frac{576\mathrm{\pi <!--\; \pi \; -->}}{35}$
C) $\frac{144\mathrm{\pi <!--\; \pi \; -->}}{35}$
D) $8\mathrm{\pi <!--\; \pi \; -->}$
I started off by writing the integral like this:
${\int <!--\; \int \; -->}_0^8\frac{1}{2}\mathrm{\pi <!--\; \pi \; -->}(\frac{1}{2}(\sqrt{8x}-<!--\; -\; -->\frac{x^2}{8}){)}^{2}dx$
then I simplified it to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}{\int <!--\; \int \; -->}_0^8(\sqrt{8x}-<!--\; -\; -->\frac{x^2}{8}{)}^{2}dx$
I can't think of anyway to solve this problem from here, without using a calculator.

Finding volume of the region enclosed by $x^2+y^2-<!--\; -\; -->6x=0,z=\sqrt{36-<!--\; -\; -->x^2-<!--\; -\; -->y^2}$
$z=0$
I'm trying to calculate the volume of the region enclosed by the cylinder $x^2+y^2-<!--\; -\; -->6x=0$, the semicircle $z=\sqrt{36-<!--\; -\; -->x^2-<!--\; -\; -->y^2}$ and the plane $z=0$. I tried moving the region towards -x so that the cylinder has it's center at zero. I ended up using these equations and proposing this integral using cylindrical coordinates:
New cylinder $x^2+y^2=9$
New semicircle $z=\sqrt{36-<!--\; -\; -->(x+3{)}^2-<!--\; -\; -->y^2}$
${\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^3{\int <!--\; \int \; -->}_0^{\sqrt{36-<!--\; -\; -->(rcos(\mathrm{\theta <!--\; \theta \; -->})+3{)}^2-<!--\; -\; -->(rsin(\mathrm{\theta <!--\; \theta \; -->}){)}^2}}r\cdot <!--\; \cdot \; -->dzdrd\mathrm{\theta <!--\; \theta \; -->}$
For some reason I can't solve it this way and I don't know what am I doing wrong.

Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross section
I'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through covering every cross section throughout the height of the 'cone' is constant. Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Finding surface area and volume of a sphere using only Pappus' Centroid Theorem
I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.
I did some calculation and found out the ratio of surface area and volume. Here is my work,
1. My key observation is finding out that the centroid of a semidisk of raidus r is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)
2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.
3. Then let's say our centroid is located h distance above the center. We still don't know it.
4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius r. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius 2r/3.
5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius 2r/3 by the factor 9/4 to get the surface area of the sphere of radius r.
Here is the calculations,
$V=2\mathrm{\pi <!--\; \pi \; -->}h\ast <!--\; \ast \; -->\mathrm{\pi <!--\; \pi \; -->}{r}^2/2={\mathrm{\pi <!--\; \pi \; -->}}^2{r}^{2}h$
$S=9/4\ast <!--\; \ast \; -->2\mathrm{\pi <!--\; \pi \; -->}h\ast <!--\; \ast \; -->\mathrm{\pi <!--\; \pi \; -->}(2r/3)=3{\mathrm{\pi <!--\; \pi \; -->}}^{2}rh$
Since we don't know h, I simply divide them to cancel it out and get, $V/S=r/3.$

When do we use $2\mathrm{\pi <!--\; \pi \; -->}$ vs using $\mathrm{\pi <!--\; \pi \; -->}$ in finding the volume of a region?
I've seen that sometimes finding the integral volume of a rotated shape in the xy-plane is multiplied by $2\mathrm{\pi <!--\; \pi \; -->}$, and other times it is only multiplied by $\mathrm{\pi <!--\; \pi \; -->}$. Can someone please tell me the difference?

Finding volume of solid in one quadrant - divide total volume by 4? 8? 2?
I want to find the volume of the solid produced by revolving the region enclosed by $y=4x$ and $y=x^3$ in the first quadrant. The wording about the first quadrant confuses me but here's my work so far:
I know the volume unrestrained by quadrant is:
$V={\int <!--\; \int \; -->}_a^b\mathrm{\pi <!--\; \pi \; -->}(f(x{)}^2-<!--\; -\; -->g(x{)}^2)dx$
Where $f(x)=4x$ and $g(x)=x^3$. To find a and b, I look for the largest and smallest intersection points between the two functions:
$\begin{array}{rl}f(x)& =g(x)\\ 4x& =x^3\\ 0& =x^3-<!--\; -\; -->4x\\ & =x(x-<!--\; -\; -->2)(x+2)\\ x& \in <!--\; \in \; -->\{-<!--\; -\; -->2,0,2\}\end{array}$
Plugging all of these into the volume equation above:
$\begin{array}{rl}V& ={\int <!--\; \int \; -->}_a^b\mathrm{\pi <!--\; \pi \; -->}(f(x{)}^2-<!--\; -\; -->g(x{)}^2)dx\\ & ={\int <!--\; \int \; -->}_{-<!--\; -\; -->2}^2\mathrm{\pi <!--\; \pi \; -->}((4x{)}^2-<!--\; -\; -->(x^3{)}^2)dx\\ & =\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_{-<!--\; -\; -->2}^2(16x^2-<!--\; -\; -->x^6)dx\\ & =\mathrm{\pi <!--\; \pi \; -->}({\int <!--\; \int \; -->}_{-<!--\; -\; -->2}^{2}16x^{2}dx-<!--\; -\; -->{\int <!--\; \int \; -->}_{-<!--\; -\; -->2}^2{x}^{6}dx)\\ & =\mathrm{\pi <!--\; \pi \; -->}({\left[\frac{16x^3}{3}\right]}_{-<!--\; -\; -->2}^2-<!--\; -\; -->{\left[\frac{x^7}{7}\right]}_{-<!--\; -\; -->2}^2)\\ & =\mathrm{\pi <!--\; \pi \; -->}(\frac{16(2{)}^3}{3}-<!--\; -\; -->\frac{16(-<!--\; -\; -->2{)}^3}{3}-<!--\; -\; -->\frac{(2{)}^7}{7}-<!--\; -\; -->\frac{(-<!--\; -\; -->2{)}^7}{7})\\ & =\mathrm{\pi <!--\; \pi \; -->}(\frac{128}{3}+\frac{128}{3}-<!--\; -\; -->\frac{128}{7}+\frac{128}{7})\\ & =\mathrm{\pi <!--\; \pi \; -->}(\frac{256}{3}-<!--\; -\; -->\frac{256}{7})\\ & =\mathrm{\pi <!--\; \pi \; -->}\left(\frac{1024}{21}\right)\\ & =\frac{1024\mathrm{\pi <!--\; \pi \; -->}}{21}\end{array}$
This is the volume for the entire function. I make an assumption that since I only want one quadrant and the function is symmetric about both the x- and y-axes, I simply divide it by four.
$\begin{array}{rl}{V}_{\text{whole}}& =\frac{1024\mathrm{\pi <!--\; \pi \; -->}}{21}\\ V_{\text{one quadrant}}& =\frac{1024\mathrm{\pi <!--\; \pi \; -->}}{21}\times <!--\; \times \; -->\frac{1}{4}\\ & =\frac{256\mathrm{\pi <!--\; \pi \; -->}}{21}\end{array}$
I have no way of verifying my results. Can my assumption be made, or there's a differing method I should be using here?
If I'm now working in 3D space, would I instead divide it by eight? But if I'm revolving around $x=0$, wouldn't the solid of revolution take four quadrants in 3D space, thus I should divide the total volume by 2?