i'm very bad in calculus

[logistic_guy]

here is the question

Why \(\displaystyle \int \frac{dA}{r} = b\ln \frac{r_2}{r_1}\)?


my attemb
i study calculus three years ago, i forgot everything. i'm studying advance engineering some of the questions use calculus
i hate calculus
when working with beams if the cross section area is rectangual i get this solution but i cant' understand how it solve it
i understand \(\displaystyle \int \frac{dA}{r} = \frac{A}{r} = b\frac{(r_2 - r_1)}{r}\) why my thinking is wrong?

 

[khansaheb]

here is the question

Why \(\displaystyle \int \frac{dA}{r} = b\ln \frac{r_2}{r_1}\)?


my attemb
i study calculus three years ago, i forgot everything. i'm studying advance engineering some of the questions use calculus
i hate calculus
when working with beams if the cross section area is rectangual i get this solution but i cant' understand how it solve it
i understand \(\displaystyle \int \frac{dA}{r} = \frac{A}{r} = b\frac{(r_2 - r_1)}{r}\) why my thinking is wrong?

That will be incorrect, because the functional relationship between A and r has not been defined.

 

[BeansNRice]

if you intend to pursue engineering and you hate calculus you're in for a bad time

 

[fresh_42]

The calculation by assuming [imath] A=b\cdot r [/imath] goes
[math]\begin{array}{lll} \displaystyle{\int_{r_1}^{r_2}\dfrac{dA}{r}}&= \displaystyle{b\cdot \int_{r_1}^{r_2}\dfrac{dr}{r}}\\[12pt] &=\displaystyle{b\cdot \left[\ln\,|r|\,\right]_{r_1}^{r_2}}\\[12pt] &=b\cdot \left(\ln\,r_2 - \ln\,r_1\right)\\[12pt] &=b\cdot \ln\dfrac{r_2}{r_1} \end{array}[/math]This is quite elementary and I even "saw" how [imath] A [/imath] has to be defined so that the equation makes sense.

I think you will need to practice a lot if you keep aiming at engineering. Problems like this should be easy. Here are some free books that are aimed at closing the gap between high school math and college math:

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[logistic_guy]

That will be incorrect, because the functional relationship between A and r has not been defined.

do you remember \(\displaystyle R\)? now think!! and think hard!!

if you intend to pursue engineering and you hate calculus you're in for a bad time

i'm an artist

The calculation by assuming [imath] A=b\cdot r [/imath] goes
[math]\begin{array}{lll} \displaystyle{\int_{r_1}^{r_2}\dfrac{dA}{r}}&= \displaystyle{b\cdot \int_{r_1}^{r_2}\dfrac{dr}{r}}\\[12pt] &=\displaystyle{b\cdot \left[\ln\,|r|\,\right]_{r_1}^{r_2}}\\[12pt] &=b\cdot \left(\ln\,r_2 - \ln\,r_1\right)\\[12pt] &=b\cdot \ln\dfrac{r_2}{r_1} \end{array}[/math]This is quite elementary and I even "saw" how [imath] A [/imath] has to be defined so that the equation makes sense.

I think you will need to practice a lot if you keep aiming at engineering. Problems like this should be easy. Here are some free books that are aimed at closing the gap between high school math and college math:

OpenStax | Free Textbooks Online with No Catch

OpenStax offers free college textbooks for all types of students, making education accessible & affordable for everyone. Browse our list of available subjects!

openstax.org

thank fresh_42 very much

i can't see this because i put the rectangle in \(\displaystyle x\) and \(\displaystyle y\) coordinate. the area \(\displaystyle dA = b\times dx\)
but how we just simply change \(\displaystyle dx\) to \(\displaystyle dr\)? do calculus allow this or it's trick you discover?

 

[khansaheb]

do you remember R\displaystyle RR?

YES ..... but R has NOTHING to do with the problem above - at least as it was presented.

The calculation by assuming A=b⋅r A=b\cdot r A=b⋅r goes

What is 'r'? What is 'b' ? What is 'A' ? ( these questions are for @logistic_guy - for him to think hard!!!)

 

[fresh_42]

do you remember \(\displaystyle R\)? now think!! and think hard!!

y
i'm an artist


thank fresh_42 very much

i can't see this because i put the rectangle in \(\displaystyle x\) and \(\displaystyle y\) coordinate. the area \(\displaystyle dA = b\times dx\)
but how we just simply change \(\displaystyle dx\) to \(\displaystyle dr\)? do calculus allow this or it's trick you discover?

Variable names are irrelevant. You can write a function as [imath] y=y(x)=f(x) [/imath] or in this case [imath] A=b\cdot r, [/imath] and more precisely [imath] A(r)=b\cdot r. [/imath] If we can simply set [imath] x=r [/imath] and [imath] A=y [/imath] then we first get [imath] A=A(r)=y=b\cdot r=b\cdot x, [/imath] then [imath] dA=dA(r)=d(b\cdot x)=b\cdot d(x), [/imath] and note that also [imath] r_1,r_2 [/imath] remain the same because we did not perform a transformation of variables, we only changed the letter, not the meaning, and finally

[math]\begin{array}{lll} \displaystyle{\int_{r=r_1}^{r=r_2}\dfrac{dA}{r}}&=\displaystyle{\int_{x=r_1}^{x=r_2}\dfrac{b\cdot dx}{x}}\\[12pt] &=\displaystyle{b\cdot \int_{x=r_1}^{x=r_2} \dfrac{1}{x}\,dx}\\[12pt] &=b \cdot \left[\ln\,|x|\,\right]_{x=r_1}^{x=r_2} \end{array}[/math]

 

[logistic_guy]

YES ..... but R has NOTHING to do with the problem above - at least as it was presented.

What is 'r'? What is 'b' ? What is 'A' ? ( these questions are for @logistic_guy - for him to think hard!!!)

\(\displaystyle R = \frac{A}{\int \frac{dA}{r}}\)

if you've a keen insight about curved beams, you'll see this instantly especially when this logisistic guy talke about beams cross section

Variable names are irrelevant. You can write a function as [imath] y=y(x)=f(x) [/imath] or in this case [imath] A=b\cdot r, [/imath] and more precisely [imath] A(r)=b\cdot r. [/imath] If we can simply set [imath] x=r [/imath] and [imath] A=y [/imath] then we first get [imath] A=A(r)=y=b\cdot r=b\cdot x, [/imath] then [imath] dA=dA(r)=d(b\cdot x)=b\cdot d(x), [/imath] and note that also [imath] r_1,r_2 [/imath] remain the same because we did not perform a transformation of variables, we only changed the letter, not the meaning, and finally

[math]\begin{array}{lll} \displaystyle{\int_{r=r_1}^{r=r_2}\dfrac{dA}{r}}&=\displaystyle{\int_{x=r_1}^{x=r_2}\dfrac{b\cdot dx}{x}}\\[12pt] &=\displaystyle{b\cdot \int_{x=r_1}^{x=r_2} \dfrac{1}{x}\,dx}\\[12pt] &=b \cdot \left[\ln\,|x|\,\right]_{x=r_1}^{x=r_2} \end{array}[/math]

thank very much

you explained it nicely

it's just me forget this tricks after years from the subject. i see the idea now

 

[fresh_42]

if you've a keen insight about curved beams, you'll see this instantly especially when this logisistic guy talke about beams cross section

The confusion was the choice of the letter [imath] r. [/imath] It is usually taken to name a radius, not a sidelength of a rectangular. And to write [imath] A=b\cdot r [/imath] is far shorter than lengthy explanations and discussions, especially when they only intend to figure out whether you know what [imath] A,b,r [/imath] stand for.

 

[logistic_guy]

if you've a keen insight about curved beams, you'll see this instantly especially when this logisistic guy talke about beams cross section

The confusion was the choice of the letter [imath] r. [/imath] It is usually taken to name a radius, not a sidelength of a rectangular. And to write [imath] A=b\cdot r [/imath] is far shorter than lengthy explanations and discussions, especially when they only intend to figure out whether you know what [imath] A,b,r [/imath] stand for.

this comment is for khan

It is usually taken to name a radius, not a sidelength of a rectangular

it's indeed a radius


look at the cross section have three parts each is rectangual, but it's taken from curv beam. this mean the radius go through it
i don't blame you if you don't understand this beam stuff, but khan is supposed to be experts in beams

if he's the expert and not seen this and you're the outsider and seen this
it make no sense. i'm confused

 

[fresh_42]

it make no sense. i'm confused

Me, too. There is no [imath] r [/imath] in the picture, no [imath] b [/imath], and the only [imath] A [/imath] is a point and not a cross-section.
And [imath] M [/imath] looks like a torque with no center or radius vector in sight. The problem I have is that I neither sit at your desk nor do I see what you see on your computer. I completely lack context. And context is what you are asked to provide. Otherwise, we could talk and each of us means something else.