How do I set up the integral to find the surface area if I only have the two shapes a sphere and a cylinder ?
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!
22559
Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions"   Is there a easy way to do this?   Maybe like FTOC II?
95876
I have 2 shapes (see below) a circle and a shape of the US. I want to remove the part of the US that falls outside the circle. How is this possible?    Thanks
123726
Geometrically speaking, it seems to me that if you have for example $y^2=8x$ revolved around the x-axis, taking the limit of the sum of $n$ surfaces of cylinders as $n$ approaches infinity should give you the surface area of that surface of revolution.  This is how the author initially derives the formula for finding the volume of solids of revolution. Take a rectangle under the curve over $\Delta x$ and revolve it around the axis to get an approximation of the volume of the solid over that interval. Add up those rectangles over $n$ changes in $x$ and take the limit as $n$ approaches infinity, which is the integral of the function that gives you the $y$ value (radius of that approximating cylinder) for each $x$ value.  Following the same principle, why wouldn't we be able to take those same cylinders, but instead of taking their volume, taking their surface area and take the limit as the number of those cylinders approaches zero?  In other words, in this case each $y$ value is given by $y = \sqrt{8x}$, which is the radius of that cylinder of height $\Delta x$ and an approximation of the surface area over that interval.  Why doesn't that work? Why do we need to deal with arc length? I don't understand why it doesn't work in this case, it seems to me that you're still getting a better and better approximation of surface area as those cylinders get smaller and smaller, eventually getting the exact surface area with the limit as their number goes to infinity.       PS: I saw this  but it's still not making sense visually/geometrically.
102178
I have coordinates of 3d triangle and I need to calculate its area. I know how to do it in 2D, but don't know how to calculate area in 3d. I have developed data as follows.  (119.91227722167969, 122.7717056274414, 39.3568115234375),  (119.8951187133789, 122.7717056274414, 39.38057327270508),  (121.11941528320312, 123.2818832397461, 38.41301345825195)
97860
I am a bit confused by this problem I have encountered:  A right circular cylindrical container with a closed top is to be constructed with a fixed surface area. Find the ratio of the height to the radius which will maximize the volume.  I know the volume to be $ \pi{r}^2h$, but I don't see what equation I should be solving for. How can I solve for the ratio?  Thanks
12488
Suppose we rotate the graph of $y = f(x)$ about the $x$-axis from $a$ to $b$.  Then (using the disk method) the volume is $$\int_a^b \pi f(x)^2 dx$$ since we approximate a little piece as a cylinder.  However, if we want to find the surface area, then we approximate it as part of a cone and the formula is $$\int_a^b 2\pi f(x)\sqrt{1+f'(x)^2} dx.$$  But if approximated it by a circle with thickness $dx$ we would get $$\int_a^b 2\pi f(x) dx.$$  So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can't.
51658
Why does the integral of the surface area of a sphere equal it's volume?   $$\int{4\pi r^2 \ \mathrm{d}r =\frac{4}{3}\pi r^3}$$  I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?
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:  $$\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}$$
82375
For me, intuitively, integral $2\pi y~dx$ make more sense. I know intuition can not be proof, but by far, most part of math I've learned does match with my intuition. So, I think this one should 'make sense' as well. Probably I didn't understand the way surface area is measured. It will be great if any one could tell me how 'integral $2\pi y~dx$' is wrong. (By the way, how to use mathematical symbols in texts?)
97061
I'm not sure how to solve this problem. Can you guys provide some input/hints?      Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ be three points in $\mathbb{R}^{2}$.      Show that the area of $\Delta ABC$ is given by   $\frac{1}{2}\left| \det(M) \right|$, where   $$M = \begin{pmatrix} 1 &amp; 1  &amp; 1\\  x_1 &amp; x_2 &amp; x_3\\  y_1 &amp; y_2 &amp; y_3 \end{pmatrix}$$
88940
Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$   Attempt: The projection of the Cylinder ( denoted $D$) on the $xy$ plane is a circle which has the equation: $x^2+y^2=ax ~~~\equiv~~~(x-a/2)^2+y^2=(a/2)^2 ~~~\equiv~~~~r=a \cos \theta$ ( In Polar Coordinates)   The circle has centre $(a/2,0)$ and radius $a/2$. Hence, the $y$ axis is a tangent to it.  Hence, the required volume $V = \int \int \int dv $  $= \int \int_D \int_{-\sqrt {a^2-x^2-y^2}} ^{\sqrt {a^2-x^2-y^2}}dz~~dx~dy$  $=2\int \int_D \sqrt {a^2-x^2-y^2}~~ dx dy$  Switching to Polar Coordinates :   $V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta=\dfrac {2}{3}a^3 \pi$  Although, the limits of $\theta$ look conceptually fine to me, my textbook uses the limits $0$ to $\pi$ and gives  the result $= \dfrac {2}{3}a^3 (\pi-\dfrac{4}{3})$.   Could someone please clarify why $-\pi/2$ to $\pi/2$ usage in my expression might be incorrect?  Thanks a lot for your help!
31210
I'm trying to find the volume between the surfaces defined by the following equations: $$ x^2 + y^2 + z^2 = b^2 $$ $$ y^2 \tan^2 a = x^2 + z^2 $$ I need to find the volume using two different methods: the first one is by spherical coordinates and the second one by polar coordinates. The problem is I get two different results: Using spherical coordinates I found: $$ x = \rho \sin \phi \cos \theta \\ z = \rho \sin \phi \sin \theta \\ y = \rho \cos \phi $$ $$ V= \int_0^{2\pi}\int_0^{a}\int_0^{b}\rho^2\sin\phi \:d\rho d\phi d\theta = \frac{2\pi}{3}b^3(-\cos(a)+1) $$ and using polar: $$ x = r \cos \theta \\ z = r \sin \theta \\ y = y $$ $$ V = \int_0^{2\pi}\int_0^{b\sin a}\int_0^{\sqrt{b^2-r^2}}r \:dy dr d\theta = -\frac{2\pi}{3}b^3(\cos^3(a)-1) $$ So I know I have made a mistake at some point but I can't find it. If anyone can help I'd appreciate it. Thanks!
44947
If there are 5 points on the surface of a sphere, then there is a closed half sphere, containing at least 4 of them.  It's in a pigeonhole list of problems. But, I think I have to use rotations in more than 1 dimension.  Regards
108481
how to find the area of any irregular shapes without dividing it into smaller regular shapes ?  Example Image:
49527
So I just started working with Blender 2.6 and the two books I have don't really discuss how to modify objects other than "here's edit mode..."    What I have is two cylinders with one slightly smaller than the other and protruding out of the larger one.  I need to "fillet" from the larger cylinder to the smaller one.  I decided to create a cone with the same diameter as the larger cylinder which then protrudes through the smaller one giving the illusion that the larger cylinder is reducing to the size of the smaller one.  Of course the problem is that the cone continues through and extends beyond the smaller cylinder.  My thought was to just cut a section off of the cone that lies inside the smaller cylinder.  Unfortunately my two books don't really cover doing anything like that.  So any suggestions would be greatly appreciated!
116062
I read a little bit about , but I imagine the problem changes drastically when trying to fill a sphere with unit cubes. For example, how many unit cubes could one fit into a sphere of radius five? Clearly the upper limit would be $\left \lfloor \frac{4}{3}\pi \cdot 5^3 \right \rfloor$, but I imagine there will be plenty more empty space than just a small fraction of a cube.