# please solve this integral of surface area

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#### flexx

##### Member level 5
please solve this integral of surface area , i tried many time but i think that i cant solve it , please help me

x = sin y 0 ≤ y ≤ pi revloved about y -axis

#### alitavakol

##### Member level 2
just write the perimeter of the shape for any value of y. then integral with dy.

A = ∫(2π×sin)dy for y=0..π. becuase 2π×sin equals the perimeter.

Here you go.

### flexx

points: 2

#### flexx

##### Member level 5
after my trial solution i find out :
s=∫2Πr .dL

=2Π ∫ sin y √1+(dx/dy)² dy ----------> L = √1+(dx/dy)² dy to find the length of the curve .

then 2Π ∫ sin y √1+cos² . dy

let let u = cos y then du = - sin y

then -2 Π ∫√1+u² .du

then i suppose sin²Θ+cos²Θ=1
then tan²Θ+1 = sec²Θ

-2 Π ∫ √ 1+tan²Θ . sec²Θ .dΘ

-2 Π ∫ sec³ Θ dΘ

if theres any explanation please can you writs some hints to let me undersatnd it ?

iam in the first grade , we didint study Ln functions yet .

ill be very thankfull if you have a look at my solution if theres any way to complete the integral .

i stopped in -2 Π ∫ sec³ Θ dΘ

is there right to divide it in to sec²Θ secΘ ?

#### steve10

##### Full Member level 3
Is this what you want?

If later you encounter something like √1+u^2, use 1+(sh(t))^2 = (ch(t))^2 beter than 1+(tg(t))^2 = (sec(t))^2.

#### flexx

##### Member level 5
thank you mr steve for your interesting in my problem .

still i dont get this :

If later you encounter something like √1+u^2, use 1+(sh(t))^2 = (ch(t))^2 beter than 1+(tg(t))^2 = (sec(t))^2.

you mean that i must substitute by 1+sin²Θ = cos ²Θ ????

i dont get what you mean ,

for your solving , i think you use u² , for each substitution we must substitute by

1+cos²Θ ?

and i dont get the ln function in your solveing

2 f (U) = u √1+u2 + ln (u+√1+u²)

what you mean by ln ? we didnt get log yet .

#### steve10

##### Full Member level 3
flexx said:
thank you mr steve for your interesting in my problem .

still i dont get this :

If later you encounter something like √1+u^2, use 1+(sh(t))^2 = (ch(t))^2 beter than 1+(tg(t))^2 = (sec(t))^2.

you mean that i must substitute by 1+sin²Θ = cos ²Θ ????

People often use different notations to represent samething. Here sh(t) =(exp(t)-exp(-t))/2 and ch(t) =(exp(t)+exp(-t))/2 are called hyperbolic functions. Sometimes, people use sinh(t) and cosh(t). There is a basic identity about them, 1+(sh(t))^2 = (ch(t))^2, which is very similar to 1+(tg(t))^2 = (sec(t))^2. But the differentiation relations are much simpler, we have
d/dt(sh(t))=ch(t)
d/dt(ch(t))=sh(t)
which are the main reasons that I suggest using sh(t) and ch(t) over tg(t) and sec(t). Notice that this is an suggestion, but NOT a must. There might be cases where you just have to use tg(t) and sec(t).

flexx said:
i dont get what you mean ,

for your solving , i think you use u² , for each substitution we must substitute by

1+cos²Θ ?
There are usually different ways to solve the same problem. For your problem, you can use "variable substitution", which is what you have been trying, or "integration by parts" which is what I wrote in that note. If you prefer "variable substitution", you set u=sh(t) in the integral. It can be integrated very easily. Keep in mind that you can always appeal to the original expression of the hyperbolic functions. For instance, if you encounter (ch(t))^2, you can always replace it by ((exp(t)+exp(-t))/2)^2.

flexx said:
and i dont get the ln function in your solveing

2 f (U) = u √1+u2 + ln (u+√1+u²)

what you mean by ln ? we didnt get log yet .

This is another way to represent the natural log, which I use "ln", while I use "log" for the base 10 log. Sorry for the confusion.

#### flexx

##### Member level 5
thank you mr steve 10 , really its so that you r such great teacher .

i tried to solve it by using variable substitution by using u = sinΘ

but iam sorry i think that iam not so smart like you , will you give me solution by using variable substitution ?

iam sure that i asked too many questions , but really its so important to me to learn from the people like you .

thanking you for your antecipation i remain

#### steve10

##### Full Member level 3
I hope I am not taking away your chance for a good exercise.

### flexx

points: 2

#### flexx

##### Member level 5
i just wanna say you are great man , thank you very much , now i ll revision the solving tips you used to solve this question.

but how can you write mathmatics symbols using Acrobat ?

regards mr Steve10

#### steve10

##### Full Member level 3
flexx said:
but how can you write mathmatics symbols using Acrobat ?
I am out of luck helping about that. Or you probably can't. I usually use "Scientific Workplace" to write and then compile as a pdf file.

#### flexx

##### Member level 5
any way you r such great man steve thank you very much , hope to contact you soon

thanking you for your interest , i remain

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