Consider the following background information:  I have a sphere that equally divided in to two hemisphere P and S. There is a plane that separate two different zone. Upper zone called A and lower zone called B. The angle $\alpha$ defined where the sphere is touching the intersection plane. So $\alpha$ larger means sphere is more in zone A and vice versa. The angle $\beta$ defines the orientation of the sphere. Schematic representation of this is below.  To calculate the area of $P$ portion of the sphere into $A$ is given by following equation $$ \mathrm{Area}_{P-A}=  r^2\int_{\theta=\frac{\pi}{2}-\beta}^\alpha  \int_{\phi=\arcsin(1/(\tan\theta \tan \beta))} ^{\pi -\arcsin(1/(\tan\theta \tan \beta)} \sin\theta\; d\theta d\phi. $$  After solving this equation we can get the following solution.  $$ \textrm{Area}_{P-A}= 2 r^2 \left\{ \cos (\alpha) \sin ^{-1}(\cot (\alpha) \cot (\beta)) -\tan ^{-1}\left(\frac{\cos (\beta)}{\sqrt{\sin ^2(\beta)-\cos ^2(\alpha)}}\right)\right\}+\pi  r^2 (1-\cos (\alpha)). $$  Note that the solution is defined as a function of $\alpha$ and $\beta$. If we change the value of $\alpha $ and $\beta$, the area of P into A will change eventually.     Now consider more complex case where $P$ and $S$ are not equal hemisphere.          Question:   If $P$ and $S$ are not equal, then what should I consider if I want to calculate the area of $P$ into $A$ as a function of $\alpha$ , $\theta$ and $\beta$? Here $\theta$ defines the distance of the cap edge from center.   Note:    I have seen this question:      where it solves for two spherical cap separate each other by an angle and calculate the area of their intersection. In my problem, I would like to calculate the area of one spherical cap when it cross a fixed intersection. I am looking for an analytical solution like the top one so that I can be able to calculate area when sphere going up and down and it rotate.
19704
Given a sphere of radius $r$ with two spherical caps on it defined by the radii ($a_1$ and $a_2$) of the bases of the spherical caps, given a separation of the two spherical caps by angle $\theta$, how do you calculate the surface area of that intersection?   To clarify, the area is that of the curved surface on the sphere defined by the intersection. At the extreme where both $a_1,a_2 = r$, we would be describing a spherical lune.   Alternatively define the spherical caps by the angles  $\Phi_1 = \arcsin(a_1/r)$ and $\Phi_2 = \arcsin(a_2/r)$.
18276
Has the below question been answered here before?  Prove: For a given enclosed volume, a sphere has minimum surface area.  Please provide link or ways to solve it. I know it is a problem of Minima and involves finding derivative and second derivatives. However, how to to frame the leading equations which has to be differentiated?  I am looking for a way to solve only using High School level Calculus. I think there must be an answer.
56624
I have a $3$-D sphere of radius $R$, centered at the origin.   $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are two points on the sphere.   The Euclidean distance is easy to calculate, but what if I were to restrict myself to traveling on the surface of the sphere?    Two approaches come to mind: use  in some way, or simply use trigonometry: calculate the angle between the two points and get a distance from that.    Will both/either of these methods work? Which would be easier?  Somewhat related to . Maybe it will inspire someone to go answer it!
56999
The rectangular equation is $$x^2+y^2-8z^2=0$$ $$x^2+y^2=8z^2$$ Know in the relationship between rectangular and spherical coords. we can manipulate our given to fit the form: $$x^2+y^2+z^2=9z^2$$ $$\rho=x^2+y^2+z^2, \space z=\rho\cos(\phi)$$ $$\rho^2=9\rho^2\cos^2(\phi)$$ $$1=9\cos(\phi)$$ $$\frac{1}{3}=\cos(\phi)$$ $$\arccos(\frac{1}{3})=1.23 \space rads$$ And so the equation in spherical coords. is $\phi=1.23$ I know my math is correct but I have the wrong answer so I'm not sure where I went wrong.
55193
Given the following image:   $\hskip{1.5 in}$   Supposing $A(100, 300)$ and $B(300, 100)$, how can I find the angle $\alpha$ between A and B?  On a side note, what's the main difference between a point and a vector? translating a point to a vector is as simple as Point = Vector? Sometimes I find articles where the terms are interchangable
119380
I know that $(x-x_0)^2+(y-y_0)^2-r^2=0$ is a general planar circle and $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2-r^2=0$ is a general sphere.  I want to know the general expression of a circle in space.  Can anyone please give me some advice here?
41572
If it is possible to express  $ \cos n \alpha $ in terms of $ \cos \alpha $  as a power series for integer $n$ ...  I like to see an expression for the quotient angle that obviously tallies when $ (\alpha , \beta) $ are swapped.  EDIT1:  Something like:  $$ \cos (\alpha - \beta)= \cos \alpha \cos \beta  + \sin \alpha  \sin \beta $$  EDIT2:  Like to know why $ \cos ( \alpha / \beta )$ cannot be expressed in terms of $ \cos \alpha, \cos \beta $, but  $ \cos ( \alpha + \beta) $  can be expressed in terms of  $\cos \alpha $ and $ \cos \beta. $
95085
I have to calculate the volume of the solid bounded by $$x^{2}+y^{2}+z^{2}=6 \qquad z=x^{2}+y^{2} \qquad z\geq0$$ using double integrals.  When I drew it, I could see that a part of the sphere is above the paraboloid. So, for me, it should be something like that: $$V=\iint_{B}\left[\sqrt{6-x^{2}-y^{2}}-\left(x^{2}+y^{2}\right)\right]dx\,dy$$ But my problem is to find these integrals. I think I'm supposed to not use spheric coordinates (since I'm calculating a double integral, not a triple one), but even the polar ones, I don't know how to find them.
55335
I was wondering how one would proceed to convert between coordinate systems in $ \mathbb R^n $.  For $ \mathbb R^2 $ the conversion is easy and just basic trigonometry. Given $(r, \theta)$ we can convert these polar coordinates to cartesian coordinates to $(x=r\cos\theta, y=r\sin\theta)$  The same follows for $ \mathbb R^3 $. Given $(r,\theta_1,\theta_2)$ in spherical coordinates we can convert the spherical coordinates to cartesian coordinates with the following transformation $(x=r\cos\theta_2\cos\theta_1,y=r\cos\theta_2\sin\theta_1, z=r\sin\theta_2)$, just so my point of reference is clear:$\theta_1$is the angle lying in the x-y plane, and  $\theta_2$ is the angle the vector makes with the x-y plane.  So what about for four dimensions and up? For the sake of simplicity we can define the n-sphere vector as $(r,\theta_1,\theta_2,...,\theta_n)$ and the cartesian vectors $(x_1,x_2,...,x_n)$ and maintain the same reference of $\theta_1,\theta_2$ as before.  How do we define a transformation in $ \mathbb R^4 $?  A look into this wikipedia page under the  gave me something that I don't believe is quite right.  As according to the section  $(x_1=r\cos\theta_1,x_2=r\sin\theta_1 \cos\theta_2,x_3=\sin\theta_1 \sin\theta_2\cos\theta_3,...)$ which is already in contradiction to my statement about $ \mathbb R^3$ transformations.   What is the general transformation rule? Am I missing something that I should know from linear algebra that I might be forgetting because my linear algebra is rusty? If so, please let me know what I should review or point me in the right direction.
3060
Don't let the simplicity of this diagram fool you. I have been wondering about this for quite some time, but I can't think of an easy/smart way of finding it.  Any ideas?    For reference, the Area is:  $$\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}$$
83676
I have the following data:-   I have two points ($P_1$, $P_2$) that lie somewhere on the ellipse's circumference. I know the angle ($\alpha$) that the major-axis subtends on x-axis. I have both the radii ($a$ and $b$) of the ellipse.   I now need to find the center of this ellipse. It is known that we can get two possible ellipses using the above data.  I have tried solving this myself but the equation becomes so complex that I always give up.  This is what I have done till now:-  I took the normal ellipse equation $x^2/a^2 + y^2/b^2 = 1$. To compensate for the rotation and translation, I replaced $x$ and $y$ by $x\cos\alpha+y\sin\alpha-h$ and $-x\sin\alpha+y\cos\alpha-k$, respectively. $h$ and $k$ are x and y location of the ellipse's center.  Using these information I ended up with the following eq:- $$a B_1\pm\sqrt{a^2 B_1^2 - C_1(b^2 h^2 - 2 A_1 b^2 h)} = a B_2\pm\sqrt{a^2 B_2^2 - C_2 (b^2 h^2 - 2 A_2 b^2 h)} \quad (1)$$  where $A = x\cos\alpha +y\sin\alpha$, $B = -x\sin\alpha+y\cos\alpha$ and $C = a^2 B^2 + A^2 b^2 - a^2 b^2$.  Now the only thing I need to get is $h$ from (1). All other values are known, but I am not able to single that out.  Anyway if the above equations looks insane then please solve it yourself, your way. I could have drifted into some very complicated path.
56821
A conical vessel with semi-vertical angle 30 degrees and height 10.5 cm has a thin lid. A sphere kept in the vessel touches the lid. The radius of the sphere in cm is ...
120080
Background:  I recently began taking calculus and it has come to alter the way I look at circles, and curves.  The equation of a circle is $\pi r^2$, traditionally in school we have always left the answer in terms of pi (i.e.) if the radius $r=2$ then the area $A = 4\pi$.    Question:  If one were to attempt to write the area of a circle in decimal form (i.e.) if the radius=2 then the area $A = 4\pi$, but $\pi$ doesn't have an end, it has (per what I have learned in school) an infinite number of decimal places so it is $3.14159\ldots$ therefore if one multiplied $4\cdot 3.14159\ldots$ one would have to approximate ones answer.    Does that mean that it is impossible to calculate the exact (without approximation) area of a circle?  Thanks for any responses,  Joel
91375
Just to check that I'm on the right track:  Are all great-circles on the sphere and in equirectangular projection (i.e. latitude, longitude pairs) either:   meridians (i.e. going pole to pole) of the form tan latitude = sin360(longitude + rotation) * amplitude + offset   (with additional restrictions on the offset/amplitude combinations -- obviously, all great circle paths with an amplitude of 0 also have offset 0 - the equator).  Or are there great-circle paths that do not fit into this scheme (again, only in a longitude-latitude coordinate system, not on other map projections).  Note: I added the tan above after posting the question, in a reaction to whubers excellent reply. It turns out that offset then always is 0.
123104
The area of ​​the portion of the sphere $ x^{2} + y^{2} +z^{2} = 1$ located inside of the cylinder $x = x^{2} + y^{2}$, and above the plane $z = 0$.  I'm stuck, so any tip will be helpful  Thanks in advance!
58033
Let's consider two spheres in the $(x,y,z)$ 3D-space, both centered in the origin: the inner with radius $r$ and the outer with radius $r + dr$.   To compute the volume of the spherical shell between their two surfaces, one should simply proceed as follows:  $$dV = \frac{4}{3} \pi (r + dr)^3 - \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r^3 + 3 r^2 dr + 3 r dr^2 + dr^3 - r^3)$$  $$dV = \frac{4}{3} \pi (3 r^2 dr + 3 r dr^2 + dr^3)$$  but usually the last two terms between the brackets are neglected. I know $dr$ is infinitesimal, but can infinitesimals (raised to $n$-th power, $n&gt;1$) be neglected and still obtain an exact result? How can this be justified?
52684
I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating   $$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$  So we should be able to find the area of the ellipse by   $$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$  $$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$  $$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$  $$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$  $$=\pi\frac{a^2+b^2}{2}.$$  First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as .  What am I doing wrong?