Why this triple integral is not null?

[Amaelle]

Homework Statement

look at the image

Relevant Equations

integrating impair function over symmetric region

Greetings
here is my integral
Compute the volume of the solid

and here is the solution (that I don't agree with)

So as you can see they started integrating sinx from 0 to pi and then multiplied everything by two! for me sin(x) is an odd function and it's integral should be 0 over symmetric region!
thank you in advance!

 

[Leo Liu]

I am confused. Definite integration is a technique to find the area under the curve. Sin(x) is completely above the x-axis between (0, Pi). So the value isn't 0.

 

[FactChecker]

By definition, volume is positive and all its pieces have positive volume. So the fact that sin() is odd forces you to consider the negative part as an additional positive volume.

 

[Orodruin]

Homework Statement:: look at the image
Relevant Equations:: integrating impair function over symmetric region

So as you can see they started integrating sinx from 0 to pi and then multiplied everything by two! for me sin(x) is an odd function and it's integral should be 0 over symmetric region!
thank you in advance!

sin(x) is odd relative to zero. Your integration domain is not symmetric relative to zero. Hence, your argument does not apply.

 

[Steve4Physics]

So as you can see they started integrating sinx from 0 to pi and then multiplied everything by two! for me sin(x) is an odd function and it's integral should be 0 over symmetric region!
thank you in advance!

The area of (say) a circle, centred somewhere on the x-axis, is not zero - even though half the area is above the x-axis and half is below.

Beware of the difference between signed and unsigned areas!

 

[Amaelle]

The area of (say) a circle, centred somewhere on the x-axis, is not zero - even though half the area is above the x-axis and half is below.

Beware of the difference between signed and unsigned areas!

thank you I got it!

 

[Amaelle]

The area of (say) a circle, centred somewhere on the x-axis, is not zero - even though half the area is above the x-axis and half is below.

Beware of the difference between signed and unsigned areas!

sin(x) is odd relative to zero. Your integration domain is not symmetric relative to zero. Hence, your argument does not apply.

thank you very much! I understood the difference now!