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the definite integrand of f(x)=cosx from -pi/2 to pi/2
sinx from -pi/2 to pi/2 = 1-(-1)=2

@lgbasallote lol jk...idk shell using trig functions

|dw:1337557715702:dw|

The height would be 2cosx, and the radius would be cosx - (-1)?

ie.
h = 2y
r = y - (-1)
?

|dw:1337558085638:dw|

Wait, I think the height would be:
h = 2x, due to symmetry

|dw:1337558202025:dw|

So then, radius is r = cos(x) - (-1) for sure, right?

And I can restate that cos(x) as y
So then, r = y +1, h = 2x = 2arccos(y), I believe.

now you are on the right track.

if you are familiar with winplot, you can make a nice 3d representation.

@shivam_bhalla! Yes you finally came.

|dw:1337558589994:dw|

x = pi/2, x = -pi/2 and y = cosx|dw:1337561871090:dw|

|dw:1337562804283:dw|

Ok so it is:
h = x
y = cosx - (-1)

I think I came up with this a few posts back

height is a function of x;
y=cos(x)
arccos(y) = x = height

|dw:1337563244465:dw|

How the heck can you resolve the radical when you have:
|dw:1337563293983:dw|
As your argument?

have you tried looking at it form both forms?
\[\sqrt{1+(x')^2}\ or\ \sqrt{1+(y')^2}\]

for the most part, i think you stick with the positive root

Why do you say "most part"

distance tends to be a measure in absolute value, so the positive root seems most apt

WOW! I just realized how EASY it is doing it wrt x.

|dw:1337563549037:dw|

Easy U-sub. Thanks a lot amistre64. What level of education is your mathematics?

I was just curious

i dunno what level it is, im guessing its at least on a 3rd grade level ;)

Ok anyways, thanks for your help

youre welcome. Levelwise is difficult since im self taught for the most part of this stuff