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We know that while deriving the surface area of cone, for differential area, we multiply the slant height element and the circumference of a small circular portion.

So when we calculate the area under curve, why don't we take $y\,\mathrm dl$ where $\mathrm dl=\mathrm dx\,\sec\alpha,$ where $\tan\alpha$ is the instanteneous slope at that $(x,y)\;?$ I think this is more appropriate since the error is not like in the case of assuming $y\,\mathrm dx$ with rectangles where some portion of area is unaccounted for.

ryang
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Paracetamol
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  • Because the area of the trapezoid is essentially the rectangle (which is $y,dx$) plus the little triangle at top (which is "on the order of $dx{}^2$" and can be ignored). – Brian Tung Jul 16 '22 at 20:51
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    As another argument against using $dl$: Consider what would happen if you were to compute the area under $y = 10x$ this way. Each trapezoid would use as its base a segment $\sqrt{101}$ times as long as it should be. You would get a result of $\left.10\sqrt{101}x\right]_{x=a}^b$ over the interval $[a, b]$. – Brian Tung Jul 16 '22 at 20:53
  • Oh then why we take the dl into account while finding surface area of cone ? @BrianTung there too its second order isnt ? – Paracetamol Jul 16 '22 at 20:55
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    Because in one case, the extra bit goes away in the limit, and in the other case, it doesn't. If you compute the area, the collective base of the little triangles at the top stays the same (the width of the interval), but the altitudes vanish in the limit, leaving zero. If you compute the length of the function's graph, however, then ignoring the little triangles leaves you with two legs of a right triangle, where you really want all the little hypotenuses. Something similar happens with the area of a surface of revolution. – Brian Tung Jul 16 '22 at 20:55
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    Can you mathematically write that as a comparison between both as answer? The area under curve and area of cone so that i can see how taking limit it goes away while in another its not the case? @BrianTung – Paracetamol Jul 16 '22 at 20:57
  • Not at the moment. Maybe later, or perhaps someone else will fill in the details... – Brian Tung Jul 16 '22 at 20:57
  • Oh yeah sure but thanks for the comments – Paracetamol Jul 16 '22 at 20:58
  • In the case of the cone, angle of any portion of slant line is the same no matter if you are at top or bottom. In the case of plain curves, they can be different. So although you can do it using $dl$ approach, it is painfully complicated with no benefit. – sku Jul 16 '22 at 21:18
  • As you say, $y,\mathrm{d}x$ is the area of the rectangle. But $y,\mathrm{d}\ell$ is NOT the area of the trapezoid, as you seem to argue. – André Caldas Jul 16 '22 at 23:08
  • Hmm understood but whats with cone or any curve surface area calcualtion why we do that ? – Paracetamol Jul 17 '22 at 01:46
  • Oh i think i understand what you were saying now , so is it true the area caluclated by the integral is exact ? Or is it that its approimxate only since that height which we say tend to zero is still very close to 0 but not exact zero? @BrianTung – Paracetamol Jul 17 '22 at 01:49
  • " however, then ignoring the little triangles leaves you with two legs of a right triangle," by two legs you meant ? The dy and dl one ? Only dx part being integrated is wrong ? You meant this right ? @BrianTung – Paracetamol Jul 17 '22 at 02:10
  • You're comparing apples and oranges. Integrating to get the area of the cone would be equivalent to integrating to get the length of the curve, which would use dl, while integrating to get the area under the curve would be equivalent to integrating to get the volume of the cone, which would use dx. – Neil Jul 17 '22 at 06:46
  • For volume why we take variable cylinders why not a frustum and integrate for volume? @Neil – Paracetamol Jul 17 '22 at 09:00
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    Roughly speaking: using $y,\mathrm dx$ in the approximation of the area under the curve, the relative error already tends to $0,$ so using $y,\mathrm dl$ is unnecessary (though not wrong). Check out this comment. – ryang Jul 17 '22 at 10:25
  • Oh i see when we use ydl what we need to subtract in order to get error to be zero ? @ryang – Paracetamol Jul 17 '22 at 11:52
  • Eh? (Please care more about punctuation and phrasing! Your posts are are more useful to future readers if they are coherent. I'm referring both to your post, which I edited, as well as to your comments here). If, by "to be zero", you mean "tends to zero", then Yes. And I found you two more Answers: 2nd, 3rd. – ryang Jul 17 '22 at 12:49

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Also known as a Riemann sum, the left hand rule version of a definite integral is

$$\int_a^b f(x)dx=\lim_{N\to\infty}\frac{b-a}N\left(f(a)+ f\left(a+\frac{b-a}N\right)+ f\left(a+2\frac{b-a}N\right)+… f\left(a+(N-1)\frac{b-a}N\right)\right)=\lim_{N\to\infty}\color{darkgray}{\frac{b-a}N\sum_{n=0}^{N-1}f\left(a+n\frac{b-a}N\right)}$$

while the formula for the trapezoid rule form of one is:

$$\int_a^b f(x)dx=\lim_{N\to\infty}\frac{b-a}{2N}\left(f(a)+2f\left(a+\frac{b-a}N\right)+2f\left(a+2\frac{b-a}N\right)+…+2f\left(a+(N+1)\frac{b-a}N\right) +f(b)\right)=\lim_{N\to\infty}\color{darkgray}{\frac{b-a}N}\left(\frac{f(a)+f(b)}2+\color{darkgray}{\sum_{n=1}^{N-1}f\left(a+n\frac{b-a}N\right)}\right)$$

where $N$ is the number of subintervals used to estimate the definite integral, like in the question’s drawing. Now compare both dark gray parts to see that the trapezoid rule is very similar to the left hand, or right hand or midpoint, rule.

Тyma Gaidash
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  • I meant to say that integral f(x) dx is what the area integral means that implies it already used rectangle of dx length . I was referring to why integral cannot be written as integralf(x) dl as definition since we do the same while calculating curved surface area isnt ? Of a solid etc – Paracetamol Jul 17 '22 at 01:40
  • And isnt there a difference in both summation the f(b) term is not there in first after taking n= 0 out isnt ? So that we left with n= 1 to N-1 that part is only same isnt? – Paracetamol Jul 17 '22 at 01:42
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The problem is that $\delta l \times y$ is not the area of the purple region in your illustration, that area is $\delta x \times y + \frac12\delta y\times\delta x$.

To see this, let's use an extreme example, the area under $y=1000 x$ from $x=0$ to $1$. If we use $y\;dl$ for the "area", the area we arrive at for the whole curve is $y\times\delta l=y\sqrt{(1000 \delta x)^2+\delta x^2}\approx y\delta x\times1000=1,000,000 \times x\delta x$, so about 500,000 units. If we use $y\delta x$ for the area we arrive at an area of about 500 units.

Which is closer to the actual area?

Suzu Hirose
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  • I understand but why in solid object surface area we say that slant length dl into that part circle cirumference is equivalent to area of rectangle of length equal to circumference and width = dl ( slant height) and not the dl component along vertical axis of symmetry of cone ? – Paracetamol Jul 17 '22 at 01:45