Before I read the formula of the area of revolution which is $\int 2\pi y \,ds$, where $ds = \sqrt{1 + \frac{dy}{dx}^2}$, I thought of deriving it myself. I tried to apply the same logic used for calculating the volume of revolution (e.g., $\int \pi y^2 dx $). My idea is to use many tiny hollow cylinders (inspired from the shell method), each has a surface area of $(2\pi y) (dx)$:  $2\pi y$ is the circumference of the cylinder, and $dx$ is the height of the cylinder  Their product is the surface area of the hollow (e.g., empty from the inside) cylinder. With this logic, the area is $\int 2\pi y dx$. Where is my mistake? Also it's confusing why for the volume it was enough to partition the object using cylinders and for areas not.
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.
15152
Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer)  I saw no example of this problem anywhere.. I saw an example how to solve it without calculus but I'm afraid that'll be deductions on my test.. I have the volume of a cylinder and the method of area between two curves but this is out of my reach.... I don't know what I can try...
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
80683
I was given a question which was like:     Suppose that a circle, square, rectangle and triangle have same perimeter. How area their areas related??   My work:  I broke the question in parts and tried to prove it seperately:  STEP $1$. Suppose the given perimeter is $P$.So the radius of circle will be ${r=P/2\pi}$.  And hence area of the circle will be:   ${P^2/4\pi}$.  Side of square will be $P/4$ and hence area of square will be $P^2/16$.   It is obvious that ${P^2/4\pi}$  >  $P^2/16$.   So Area of circle is more than area of square.  STEP $2$: Whether by using $AM&gt;GM$ inequality or by using a bit of differentiation, we can get that product off two quantities is maximum when they are equal.  So, Area of square is more than Area of rectangle.  Here, the hardest part come (at least for me). I can't relate the area of triangle to any of Circle, Square or Rectangle. It would have been easy if the triangle was Equilateral, Right angle but the question is talking about a general triangle.  However, when I considered the situation with different examples (Assuming the perimeter and calculating the area, I got the relation:  Circle > Square > Rectangle > Triangle.  I shall be highly thankful if you guys can establish the relation in purely mathematical way.  EDIT: This question does not ask that whether circle have the largest area among all 2-D figures or not. The OP just want to relate the areas of Circle, Square, Rectangle, Triangle in a pure mathematical way.
79877
I have this difficult integral to solve.  $$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$  Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!  Edit:  This is what I did (showing just for the numerator)  $$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$ $$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx  $$ $$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx  $$  (taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)  $$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx  $$  Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
61730
A hole of radius r is bored through the center of a sphere of radius $R &gt; r$. Find the volume of the remaining portion of the sphere.    I'm not sure on how to approach this problem. What is the most difficult part is knowing how to set up the bounds of integration. Mainly, how does one set up the bounds either when regarding cylindrical shells or washers? How do we exactly arrive at this (is it the points in which we do the sum of the tiny elements of area $dA$)? Is there a "fool-proof" way of knowing the bounds?  I was trying to solve this problem using the washer method in terms of y, but I'm not sure on how to set up my bounds of integration.  Thank you very much for your help in advance.
7891
I've seen how to visualize the formula for getting the area of a parallelogram. The first picture shows 2 ways which give the same result of Area = base * height.   (link to first pic)  However, the first time I thought about it before being show the actual formula, I thought it was Area = side * side.   (link to second pic)  The 2 sides of the parallelogram are 2 vectors, x and y.  If you place copies of vector y length of x times along the vector x, and then you add all lengths of vector y's (which is the same as length of vector y * length of vector x), it seems like you also get the area of the parallelogram.   I know that you'll end up with a different result since the height is always less than the length of the vector y (unless they were parallel) but the method in the second picture seems correct.   I also haven't seen the second picture's method being taught online so I'm guessing it's actually incorrect. What I want to know is what makes it incorrect because it really seems like it's correct despite giving a different answer from the usual base * height formula.  Thanks in advance for any help.
105314
I need to calculate the volume of solid enclosed by the surface $(x^2+y^2+z^2)^2=x$, using only spherical coordinates.  My attempt: by changing coordinates to spherical: $x=r\sin\phi\cos\theta~,~y=r\sin\phi\sin\theta~,~z=r\cos\phi$ we obtain the Jacobian $J=r^2\sin\phi$. When $\phi$ and $\theta$ are fixed, $r$ varies from $0$ to $\sqrt[3]{\sin\phi\cos\theta}$ (because $r^4=r\sin\phi\cos\theta$). Keeping $\theta$ fixed, we let $\phi$ vary from $0$ to $\pi$. Thus the volume equals:   $$V=\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\sqrt[3]{\sin\phi\cos\theta}}r^2\sin\phi ~dr ~d\phi ~d\theta=0$$ Which is obviously wrong. What am I doing wrong?
103061
I came across this question from Schaum's Outlines of Advanced Calculus:     Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x$, $y^2=8x$, $x^2=y$, $x^2=8y$ about the $x$-axis is $\dfrac{279\pi}{2}$.   My solution was to divide the region into three parts:    $$\pi\int_{1}^{2}(x^2)^2 - (\sqrt{x})^2\;dx = 4.7\pi$$  $$\pi\int_{4}^{8}(\sqrt{8x})^2 - (\sqrt{x})^2\;dx = 42\pi$$  $$\pi\int_{2}^{4} (x^2)^2 - \left(\frac{1}{8}x^2 \right)^2\;dx = 44.9\pi.$$  Hence the total volume will be  $$\pi(4.7 + 42 + 44.9) = 91.6\pi$$  which is wrong.  Can someone point out my mistake?
68115
If you take the derivative of the volume of a sphere, $$\frac{4}{3}\pi r^3$$you get its surface area, $$4\pi r^2.$$ If you differentiate again, you get $$8 \pi r.$$ Does this have any physical (or other kind of) significance, besides being 4 times the length of a great circle on the sphere?
56956
I wondered if the centroid of an area could be derived as a limit from the centroid of the solid of revolution built by the same area revolving around an axis of revolution.   e.g.: the  is $C_{V} = \frac{3\pi}{16}r\frac{\sin[\theta]}{\theta}$, where $r$ is the radius of the sphere and $\theta$ is the dihedral angle. The  is $C_{A} = \frac{4r}{3\pi}$  Now I tought it was possible a limit procedure like $\lim\limits_{\theta\to 0} C_{V}$=$C_{A}$, but that's false and I would like to know why (I guess that has something to do with the passage from three-dimension to two-dimension)  Thanks
86152
The classic proof for $\frac{\sin x}x\to1,x\to0$ is using a squeezing theorem based on arguments about areas of circles.   But as far as I know, all deduction of formula of circles' area is based on the fact $\frac{\sin x}x\to1,x\to0$. For example, using definition of definite calculus needs the fact to show limitation; using Newton-Leibniz formula needs to know $(\sin x)'=\cos x$, whose deduction also needs the fact $\frac{\sin x}x\to1,x\to0$.  So is it a circular reasoning?
27512
This problem occurred to me when I came across a similar problem where the radii were taken over only the primes. That question was unanswered, but it seems to me infinitely many circles of radius $1/2, 1/3, 1/4...$ can fit into a unit disk. The area of all those circles would be $\pi \sum_2^\infty 1/n^2 = \pi^3/6 -\pi$, which is less than the area of the unit disk $\pi$. But can the circles actually be packed with no overlaps?
117638
In the following diagram, let the side length of the large square be $s$.    How do I specifically determine the numerical value of the area of the blue region?  I can easily show the equivalence between the blue and yellow area, but I don't think this helps determine the area. If possible, I would like to have a method without integration.
62062
Let $f$ have power series $f(z) = \sum_{n=1}^\infty a_n z^n$ in $D$, then prove that   $\mathrm{area}\, f(D) = \sum_{n=1}^\infty n \,|a_n|^2$.  Note: We define $\mathrm{area}\, S = \iint_S \mathrm{d}x\,\mathrm{d}y$.  I presume the way to do this is to take the integral $$\iint_{f(D)} \mathrm{d}x\,\mathrm{d}y = \iint_D \mathbf{J}_f (x+iy) \,\mathrm{d}x\,\mathrm{d}y = \int_0^1 \int_0^{2\pi} r \mathbf{J}_f (r e^{i \theta}) \mathrm{d}\theta\,\mathrm{d}r,$$ and letting $\gamma_r : [0,2\pi]\to\Bbb{C}, \,\gamma_r (\theta) = r e^{i\theta},\, \gamma_r '(\theta) = i\gamma_r (\theta),$ then we get $$\mathrm{area}\,f(D) = \int_0^1 \int_{\gamma_r}\frac{\lvert f'(z)\rvert^2 \bar{z}}{ir}\mathrm{d}z\,\mathrm{d}r.$$ Unfortunately, this approach seems to be a dead end. I think I'm meant to use Cauchy's Formula somewhere, but I can't see where that might be useful in this kind of question.
52820
The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume?   Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem?   Please help, thanks!
34454
I checked this  but didn't fully understand it.   I know that the volume of a right circular cone is $V = \frac{1}{3}\pi x^2h$  I know that I must take the first derivative and set it equal to zero, which will find the maximum.   My problem is how to deal with the variable $h$, the height?   How can I rewrite that in terms of $x$ or $r$.      I have tried to implicitly differentiate this equation, but that wasn't helpful for me.