Expert Help with High School Geometry

Finding volume of a sphere
I am stuck on the following problem:
A circle $x^2+y^2=a^2$ is rotated around the y-axis to form a solid sphere of radius a. A plane perpendicular to the y-axis at $y=\frac{a}{2}$ cuts off a spherical cap from the sphere. What fraction of the total volume of the sphere is contained in the cap?
So far I have figured out the following:
Rotating the cap on the y axis we get a height h starting from $y=\frac{a}{2}$. The interval from $y=0$ to $y=\frac{a}{2}$ (the region below the cap) should be:
$a-<!--\; -\; -->h$
I also know that the radius of the sliced disk, x, can be derived from the equation of the circle:
$x=\sqrt{a^2-<!--\; -\; -->y^2}$
Since the area of a circle is $A=\mathrm{\pi <!--\; \pi \; -->}{r}^2$ the area with respect to y for the circle should be:
$A(y)=\mathrm{\pi <!--\; \pi \; -->}(a^2-<!--\; -\; -->y^2)$
So to find the volume, we need to integrate the function:
$V={\int <!--\; \int \; -->}_{\frac{a}{2}}^a\mathrm{\pi <!--\; \pi \; -->}(a^2-<!--\; -\; -->y^2)dy$
I know where I should go, but I am not sure what to do about the constraint $y=\frac{a}{2}$ at this point. Should I integrate the terms with respect to y first and then plug in the value which is equal to y? Or should this be done before integrating?

Probability with elements of geometry
We have a wire of length 20. We bend this wire in random point. And then bend again to get rectangual frame. What's the probability that area of this rectangle is less than 21.

Distance between (-7,-3,6) and (-3,5,1)?

Distance between (5,-6,4) and (-10,-2,2)?

Problem with finding the volume with cylindrical coordinates.
I have the following function:
$Z=x^2+y^2,Z=x+y$
that i want to solve (to find the volume) with cylindrical coordinates. I am evaluating the integral to get:
$V=\iiint <!--\; \iiint \; -->dV=\underset{R}{\iint <!--\; \iint \; -->}{\int <!--\; \int \; -->}_{x^2+y^2}^{x+y}dzdA=\underset{R}{\iint <!--\; \iint \; -->}x+y-<!--\; -\; -->(x^2+y^2)dA.$
and from here i am trying to get the bounds for r by intersecting the two functions and i get that $r=0orr=1$ therefore i tought that $0\le <!--\; \le \; -->r\le <!--\; \le \; -->1$ but it dosen't seems right. But i get the following result:
$2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^1(r-<!--\; -\; -->r^2)rdrd\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}(\frac{r^3}{3}-<!--\; -\; -->\frac{r^4}{4}){|}_0^1=\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$
and this is not the corrent answer, i should get $\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}.$. I know this is too specific to a given problem question i apologize for that but can someone tells me where i made a mistake

Distance between (3,2,-1) and (5,-1,6)?

Geometric series with probability
Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that John wins, and the probability M that Mary wins. Check that $M+J=1$. b.) Now John , Mary, Paul draw alternately from a deck of k cards. Calculate separately their respective probabilities of winning, given that John draws first and Mary draw second. Check that $J+M+P=1$

A Diminishing Geometric Distribution
A standard geometric distribution can be interpreted as the number of Bernoulli trials required to get one success. However, what if the probability of success if each trial diminishes by some factor with each failure?
Let p be probability of success of the first trial and d be the diminishing factor of each failure such that the probability of success of trial n is $p{d}^{n-<!--\; -\; -->1}$. I have been trying to calculate the expected value of this distribution without success. My main stumbling block is the probability of failing n times:
$P(n\text{ failures})=\underset{k=1}{\overset{n}{\prod <!--\; \prod \; -->}}(1-<!--\; -\; -->p{d}^{k-<!--\; -\; -->1})$
Given that, the expected value would be
$E(X)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}np{d}^{n-<!--\; -\; -->1}F(n-<!--\; -\; -->1)$
Is there a way to get a closed form for this?

Distance between (5,-11,-5) and (4,-13,6)?

In the following figure, $AD=AB$. Also $\mathrm{\angle <!--\; \angle \; -->}DAB=\mathrm{\angle <!--\; \angle \; -->}DCB=\mathrm{\angle <!--\; \angle \; -->}AEC={90}^{\circ <!--\; \circ \; -->}$ and $AE=5$. Find the area of quadrilateral ABCD.

$\sqrt{(x^2-<!--\; -\; -->25)}$

Proof of the Exterior Angle Theorem.
The exterior angle theorem for triangles states that the sum of “The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. The sum of the measures of the three exterior angles (one for each vertex) of any triangle is 360 degrees.” How can we prove this theorem?

So, for the triangle above, we need to prove why $\mathrm{\angle <!--\; \angle \; -->}CBD=\mathrm{\angle <!--\; \angle \; -->}BAC+\mathrm{\angle <!--\; \angle \; -->}BCA$

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

The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by $v=v_0(\frac{1-<!--\; -\; -->y}{\mathrm{\alpha <!--\; \alpha \; -->}})$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from (0,0) to $(2X,0)$ is the arc of a circle centeblack at $(X,\mathrm{\alpha <!--\; \alpha \; -->})$. Note that the origin is chosen to be well above the ground, so that $y<0$ is allowable.

Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: ${\displaystyle \frac{1}{v_0(\frac{1+y}{\mathrm{\alpha <!--\; \alpha \; -->}})\sqrt{1+y^2}}}=-<!--\; -\; -->C$ where $C$ is a constant.

Finding the volume bounded by surface in spherical coordinates
$R=4-<!--\; -\; -->1\cos <!--\; \; -->(\mathrm{\varphi <!--\; \varphi \; -->})$

3-point' curve
If you have a loop of string, a fixed point and a pencil, and stretch the string as much as possible, you draw a circle. With 2 fixed points you draw an ellipse. What do you draw with 3 fixed points?

Find the probability that a geometric random variable X is an even number
Let $\mathrm{\alpha <!--\; \alpha \; -->}$ be the probability that a geometric random variable X with parameter p is an even number
a) Find $\mathrm{\alpha <!--\; \alpha \; -->}$ using the identity $\mathrm{\alpha <!--\; \alpha \; -->}=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}P[X=2i]$.
b)Find $\mathrm{\alpha <!--\; \alpha \; -->}$ by conditioning on wether $X=1$ or $X>1$

Distance between (-6,3,4) and (-2,2,6)?

Geometric Probability Distribution, Expected Values
Let $X\sim <!--\; \sim \; -->$ Geometric $(\mathrm{\theta <!--\; \theta \; -->})$, and let $Y=min(X,100)$. Compute (a) E(Y) and (b) $E(Y-<!--\; -\; -->X)$.
I know that the Geometric distribution is $(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^{k-<!--\; -\; -->1}\mathrm{\theta <!--\; \theta \; -->}$ and I also know how to calculate expected value but I'm confused about what it means that $Y=min(X,100)$?

Suppose $ABCD$ is a cyclic quadrilateral and $P$ is the intersection of the lines determined by $AB$ and $CD$. Show that $PA·PB=PD·PC$

Using Differentials to Calculate the Volume of a Square Pyramid
Use differentials to solve the problem:
The Louvre Pyramid is a tourist attraction in Europe. It is a square pyramid, with a height of 21m, and base of side length 35m. The four faces of this pyramid are covered in glass, of thickness 0.03m. Find the volume of glass used to construct the exterior of the Louvre.
I know that the volume of a square pyramid is: $V=\frac{a^{2}h}{3}$, where a is its base length and h is its height.
I then solved for $dV=\frac{a^{2}dh+2ahda}{3}$, but I am stuck in this equation because there are many unknowns.
The next step I can think of is finding the volume, which can be in the form of: $V_{Glass}=V(Value+0.03)-<!--\; -\; -->V(Value)$, which can be evaluated like: $L(x)=f(x_0)+f^{\prime }(x_0)(x-<!--\; -\; -->x_0)$, but I am not sure of the way of finding the V(Value).