I have a problem understanding the definition of the surface area integral of a revolution . I know that we can obtain its volume using the disc method . The method defines the width of the disc as infinitely small (dx) , it is then pretty obvious that the integral is defined as the area made by each point multiplied by the width , which is dx .I found however by the definition of the surface area integral that the width is not defined as dx . Can somebody please clarify why it doesn't work the same way with surface areas ? !  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.
8860
The integration is generally area under the curve in $\Bbb R^2 \to \Bbb R$  Is integration in the range from $a$ to $b$ the same as $b$ to $a$ or is it negative?  If it is negative, Is it merely notion of convention or is there some intuition for it?
17254
When we were in school they told us that the Surface Area of a sphere = $4\pi r^2$  Now, when I try to derive it using only high school level mathematics, I am unable to do so. Please help.
28856
I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.  However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?
73606
$\int_{-2}^0\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} dydx$  y describes the circle of radius 2, and is positive and negative, so doesnt limit any quadrants in our integral, however x can only be negative, which means the left half of our circle is what we are interested in.  This means, we need the integral:  $\int_{\pi/2}^{3\pi/2}\int_{0}^{2} rdrd\theta$  Giving us a value of $2\pi$.  Is this a correct assessment? I wasnt sure if making  $0$ and $2$ the limits of $r$ was entirely correct, since it describes the entire circle of radius $2$, and not just one half. Then again, a circle of radius $2$ does have an area of $4\pi$ which makes me feel I am along the right lines.   Any clarification would be fantastic. Thank you!
65067
I am having difficulties understanding and solving the following problem.    So far I think I understand the first line. I know what the graph of floor[t] looks like and I understand they're asking to integrate from 0 to some value of n and that integral always equals the equation on the right. I do not really understand what they are asking for in the second line or how that differs from the first equation.  I have checked through my notes and various online sources and we've just covered the fundamental theorem of calculus and discussed integration using partitions on an interval to calculate area but I am also unsure about how to relate these to this problem if they are even needed.  Any pointers, explanations or solutions would be greatly appreciated.  Thank you.
71477
How do I find the volume when revolving the region bounded by $y=1-\frac{1}{2}x$, $y=0$, and $x=0$ about the line $ x=-2$?  Would it be $x=2-2y$ so radius $r(y) = 2-2y -(-2) $ => $r(y)= 4-2y$   $π\int (4-2y)^2 dy$ ?  What would be my limits of integration? Would it be from 0 to 2?? yes?
57468
So last night, my D&amp;D group ran into an interesting question. The spell Entangle says:     Area: plants in a 40-ft radius spread.    We (like a lot of people, I think) play D&amp;D on a square grid, where each square represents a 5'x5' area. In that context, how do you find the area when you're given a radius?   The books seem unclear, probably since the definition of a radius is fairly simple. My suggestion was to treat it the way you treat range: start at the center, and any square you can get to with 8 five-foot 'steps' is affected. Someone else thought it should be a square that was 80'x80'; that is, a 'circle' that's 80 feet across, since a lot of other 'circles' in D&amp;D are actually treated as squares. That's a simply huge area, though -- bigger than the board we use. We wound up using a 40x40 square, since my counting would take more time and the 80x80 square was huge. 40x40 was still big enough to encompass the entire fight.   My question: how does a radius measurement (and by extension, circles in general) work on a square grid?
89637
I know it is defined as $ dP \wedge dx  $ or explicitly,  $$ \frac{\partial P }{\partial y } dy \wedge dx   . $$  The question is, could it be $dx \wedge d P $?  Or  $$  \frac{\partial P }{\partial y }  dx \wedge dy ?    $$  I know the question must be very naive or even stupid. But I am indeed confused.
122438
Im trying to get my head around complex integration/complex line integrals. Real integration can be thought of as the area under a curve or the opposite of differentiation. Thinking of it geometrically as the area under a curve or the volume under a surface in 3 dimension is very intuitive.   So is there a geometric way of thinking about complex integration? Or should I just be viewing it as process that reverses differentiation? Or has integration other meanings in complex analysis? Here is an example, could someone explain this to me - Here's the definition of the integral along a curve gamma in C, parameterized by $w:[a, b] -&gt;C$ \begin{equation} \int_\gamma f(z)dz = \int_a^b f(w(t)).w{(t)}'dt \end{equation}   So I have -  $\gamma$ is the unit circle with anti-clockwise orientation parameterized by  $w:[0, 2\pi]\to C$  $w(t) = e ^{it} = Cos(t) + iSin(t)$  So if use the definition of the integral,   $\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}&#39;dt$,  and work this out it comes to  $\int^{2\pi}_0 i dt = 2{\pi}i$  So what does this $2{\pi}i$ represent? Does it mean anything geometrically, like if a regular integral works out to be 10 that means the area under the curve between 2 points is 10...
122758
I tried looking it up but many websites just state it without proof and without intuition.  I'm hoping to learn it a little bit better so that I don't forget how to compute the Jacobian when working with surface integrals where the divergence theorem is not applicable.  If you have a good online reference instead, please feel free to provide it :-)  Thanks,
82174
I'm looking at the Riemann surface of $f(z) = z^{1/2}$ so the set $\{(z,w) \in \mathbb{C}^2 : w^2 = z \}$. I understand that the point of the riemann surface is to understand this multi-valued function.  Now I visualise this as taking two copies of the complex plane both with a slit in them (negative real axis say) and then I put them on top of each other, flip the top one and then sort of join them up along the slit and that's fine and I can see that that will form some surface where the local coordinates are given by projection onto the z or w axis. But here I am just dealing with the domain of the function $f(z)$.  My difficulty is when I think of the analogous situation with say $\{(x,y) \in \mathbb{R}^2 : y^2 = x\}$ this "surface" is just a line and I can see how given a point on this "line" we have an associated pair $(x,y)$. Now when I look at the Riemann surface described above all I see is the domain of the function $f(z)$ I can't see how each point of the surface is in anyway related to a pair $(z,w)$.  I realise this is quite a vague question and I am struggling to put my frustration with this concept into words - I hope this makes sense! I've now added a (terrible) diagram which may help to illustrate my issue:    Thanks
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
78797
$$\det(A^T) = \det(A)$$ Using the geometric definition of the determinant as the area spanned by the columns, could someone give a geometric interpretation of the property?
71338
Since I started more or less formally learning the foundations of calculus, I naturally came across the Riemann definition of the integral. At first I considered this definition intuitive but not really useful since to obtain a closed form expression, one needed to add an infinite amount of values. Later on, an exercise prompted me to calculate a Riemann Integral, and from the definition and the expression for the sum of squares, I was able to calculate the limit with nothing more than I had learned at school.  This was a revelation for me, since so far I had considered the definition a mere formalism. Now I knew how it gave results. The next integral I tried to calculate this way was, for obvious reasons $\sqrt {1-x^2}$. Unfortunately, I found the sum intractable and gave up.  I started to question the usefulness of the definition again. If it only works for simple functions like polynomials, how did we ever find out that the integral of $\sin (x)$ is $-\cos (x)$? Did we use the Riemann definition or did we just say "the derivative of $-\cos (x)$ is $\sin (x)$ and therefore its integral must be $-\cos (x)$?  I would like to get some insight into the theory as well as the history that led to the tables of integrals we have today
99747
To find the area in a polar curve, we use areas of sectors of tiny pie slices to figure out the area, to get $$A=\int_a^b\frac{1}{2}r^2\,d\theta$$ Why can't we use the circumference of the circle to figure out arc length of a polar curve? I tried and it went something like this: The arc length of a tiny pie slice is $r\theta$, where theta is in degrees and r is the radius. Adding up all the tiny pie slice arc lengths, we should get $$\int_a^b r\,d\theta$$ But that is clearly not correct! For example, for $1+cos\theta$, this 'formula' tells us that the arc length is $2\pi$, when the actual arc length is 8. What am I doing wrong? Am I assuming something wrong? Or is it just stupid mistakes?
28612
I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?