last day (15 days later) » 

MadSpaceMemer
MadSpaceMemer
01:10
Hey . Can I ask you a quick question about integration problem I am having
Rithaniel
Rithaniel
You can, but I can't promise I'll be able to provide an answer
MadSpaceMemer
MadSpaceMemer
I am studying physics and math but i have not reached this topic in math. But i am forced to do it in physics so i kinda have background but i really dont have much infromation about it since i didnt study it in depth

I have homework to calculate the area of the following picture.

I have done this using two ways 1) with trigenometry
2) with general integral of whole space and just removing the parts that i do not need

i get a correct result of approx 0.8 each time.

However in the homework, i think it wants me to do it in a spesific way, which i am not aware of.
user image
Rithaniel
Rithaniel
Probable wants you to split it into two integrals
MadSpaceMemer
MadSpaceMemer
Well yes i thought about that but which two?
The only one i know would be taking the integra of y=2 from 0.5 to 1 then removing the
space under it by subtracting the other integrals and same for the second part.
I thought theres some technic where one can use some kind of Y and X integration in both direction
with regard to the function as whole? Or is my original way correct?
Rithaniel
Rithaniel
So you integrate y from 1/x to 2 and then integrate x from 1/2 to 1. Then the other double integral is y from x to 2 and then x from 1 to 2
Basically that chops it in half along the vertical line x=1
MadSpaceMemer
MadSpaceMemer
01:26
I understand what you mean. But why do you set the lower boundary of the integral to be "X" and "1/x'? why not "1"?
Rithaniel
Rithaniel
Because the bottom isn't a horizontal line
It's a curved shape, determined by those equations
MadSpaceMemer
MadSpaceMemer
Can I understand the lower parameter of the integral as the place from which my "rectangles" start? "considering the principle of integration
as I mentioned. I did not study this in depth
Rithaniel
Rithaniel
Gotcha. Basically you can think about your lower bound as a lower edge of whatever shape you are trying to find the area of. It's not necessarily a rectangle
MadSpaceMemer
MadSpaceMemer
02:02
With reactingles i meant building an integral is usually building infinite small reactingles to build the shape of the area to calculate, usually they start from x axes but now i understand u choose the function as lower axis.
Much thanks now i can procced. i really owe you! ;3
Rithaniel
Rithaniel
02:51
No problem. Any time
 
9 hours later…
MadSpaceMemer
MadSpaceMemer
11:28
Hey i wanted just to add something if i was say calculating the volume of some object on this area with equation z équal to something. Can i just multiply this equation to both parts of my integration ( first integral and second one ) and then derive and hold one as constant while deriving

last day (15 days later) »