Expert Help with High School Geometry

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.

You have two intersecting quadrilaterals (the area of intersection is the grey polygon with thick boundary):

These properties holds:
One quadrilateral is always a rectangle
There is always some intersection
Both quadrilaterals are convex (hence the intersection is a convex polygon as well)
The goal is to measure area of intersection (the actual shape is not needed, only a scalar showing how much space is covered by the intersection).

Distance between (-2,1,3) and (-4,0,2)?

In $\mathrm{△<!--\; △\; -->}ABC$, side BC is divided by D in a ratio of 5 to 2 and BA is divided by E in a ratio of 3 to 4 as shown in Figure. Find the ration in which F divides the cevians AD and CE i.e. find $EF:FC$ and $DF:FA$.

Distance between (-2,1,3) and (-2,0,1)?

Let $\mathrm{△<!--\; △\; -->}ABC$ be a triangle of sides AB, AC and BC which has the following equations
$$\begin{gathered} AB\equiv <!--\; \equiv \; -->a_{1}x+b_{1}y+c_1=0 \\ AC\equiv <!--\; \equiv \; -->a_{2}x+b_{2}y+c_2=0 \\ BC\equiv <!--\; \equiv \; -->px+qy+r=0 \end{gathered}$$
Then an angle $\mathrm{\angle <!--\; \angle \; -->}A$ is acute if
$\left|\begin{array}{cc}{a}_1& b_1\\ p& q\end{array}\right|\left|\begin{array}{cc}p& q\\ a_2& b_2\end{array}\right|(a_1{a}_2+b_1{b}_2)<0$
And the angle $\mathrm{\angle <!--\; \angle \; -->}A$ is obtuse if
$\left|\begin{array}{cc}{a}_1& b_1\\ p& q\end{array}\right|\left|\begin{array}{cc}p& q\\ a_2& b_2\end{array}\right|(a_1{a}_2+b_1{b}_2)>0$
But I absolutely have no idea how to prove this theorem.

Memoryless property and geometric distribution
Suppose X is a random variable taking values in ${\mathbb{N}}_0$ with the memoryless property,i.e., for each pair of number $s,t\in <!--\; \in \; -->\mathbb{N}$,
$P(X\ge <!--\; \ge \; -->s+t\mid <!--\; \mid \; -->X>t)=P(X\ge <!--\; \ge \; -->s)$
Show that a random variable X with values in ${\mathbb{N}}_0$ has the memoryless property if and only if $X\sim <!--\; \sim \; -->\text{Geometric}(p)$ of parameter $p=P(X=1)$.

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.

Distance between (-3,-2) and (-6,-7)?

Justify the small angle approximation for tangent
Assume that angle x is small, using small angle approximation,
$\sin <!--\; \; -->(x)=x$
$\cos <!--\; \; -->(x)=1-<!--\; -\; -->\frac{x^2}{2}$;
and $\tan <!--\; \; -->(x)=x$.
I am able to justify the first two using Maclaurin's Theorem but not the last one. How do we justify the last one?

How to find a probability that sum of geometric variables is less than a number
Let $X_i,i=1,\dots <!--\; \dots \; -->,n$ be Geometric i.i.d random variables, which represent the number of fails, with parameter p.
Calculate or estimate from above and below:
$P(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i\le <!--\; \le \; -->A),A\in <!--\; \in \; -->N.$
I know that sum of the geometric random variables is the negative binomial, but I would not know all the parameters for the negative binomial r.v.

How do you find all points that have an x -coordinate of –4 and whose distance from point (4, 2) is 10?

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, $V/S=r/3.$

Let quadrilateral ABCD satisfy $\mathrm{\angle <!--\; \angle \; -->}BAC=\mathrm{\angle <!--\; \angle \; -->}CAD=2\mathrm{\angle <!--\; \angle \; -->}ACD={40}^{\circ <!--\; \circ \; -->}$ and $\mathrm{\angle <!--\; \angle \; -->}ACB={70}^{\circ <!--\; \circ \; -->}$. Find $\mathrm{\angle <!--\; \angle \; -->}ADB$.

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?

Prove that a 'tangential' quadrilateral (i.e. one with an in-circle) whose area is given by Brahmagupta's formula for a cyclic quadrilateral is also cyclic (and thus 'bicentric').

I have a circle with an arc beginning at point (x,y). The radius is r, the arc angle(w/ respect to center) is $\mathrm{\theta <!--\; \theta \; -->}$. How do I calculate the end point of the arc (a,b) ?
I know that the arc-length=radius*(arc angle)
I can't seem to find an easy way to solve this, I think the way to go is with parametric equations but I'm not sure.

I would like to maximize the function:
$\frac{1}{2}\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}|<!--\; |\; -->x_i-<!--\; -\; -->\frac{1}{N}|<!--\; |\; -->$
under the constrains $\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{x}_i=1$ and $\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->(1,...,N)$
I have done some test for small values of $N$ and I have the feeling that the solution is $1-<!--\; -\; -->\frac{1}{N}$ but I can't figure out how to solve it analytically.

Distance between (8,6,2) and (4,3,1)?