Calculating the flux of a vector field

[Slimy0233]

Homework Statement

Q.18. The flux of the vector field ##\vec{F}=(x+y) \hat{i}+(x-3 y) \hat{j}+x y \hat{k}## through the surface defined by the equation ##x^{2}+y^{2} \leq a^{2}, x>0, y>0##, will be
(a) ##a^{4} / 8##
(b) ##a^{4} / 4##
(c) ##a^{2} / 8##
(d) ##a^{2} / 4##

Relevant Equations

##\vec{F}=(x+y) \hat{i}+(x-3 y) \hat{j}+x y \hat{k}##

I am not sure why latex is not rendering, but here is the question.

The answer is ##\frac{a^2}{8}## and for the love of my life, I don't know how. Can you please help me with this?

 

[anuttarasammyak]

[tex]\int_S xy dS= \int_0^a r^2 rdr \int_0^\frac{\pi}{2} \sin\theta \cos \theta d\theta[/tex]
Azimuthal integration gives 1/2.

 

[Slimy0233]

[tex]\int_S xy dS= \int_0^a r^2 rdr \int_0^\frac{\pi}{2} \sin\theta \cos \theta d\theta[/tex]
Azimuthal integration gives 1/2.

I am DOUBT! azimuthal integration is a new phrase for me. I think it's the equivalent of what I learnt as integration using spherical co-ordinates, but still, I don't see how you can get 1/2 there. Also, isn't dr 0? We are integrating with radius r as constant (a). We are given a surface [tex]x^2 + y^2 \leq a^2[/tex]

 

[Steve4Physics]

I think$$\int_{x=0}^a \int_{y=0}^a xy ~dx~dy$$should actually be$$\int_{x=0}^a \int_{y=0}^{\sqrt {a^2-x^2}} xy~ dy~dx$$
And on 'dimensional' grounds, I think the answer should be a multiple of ##a^4## rather than ##a^2##. So the 'official answer' might be wrong.

 

[Slimy0233]

I think$$\int_{x=0}^a \int_{y=0}^a xy ~dx~dy$$should actually be$$\int_{x=0}^a \int_{y=0}^{\sqrt {a^2-x^2}} xy~ dy~dx$$
And on 'dimensional' grounds, I think the answer should be a multiple of ##a^4## rather than ##a^2##. So the 'official answer' might be wrong.

can you please explain the limits of the integral more? I mean, I put a as the limit of the integral as it's a part of a circle we are talking about here, why are you putting a different limit?

edit: Damn! Yes! It's a circle we are talking about, I seem to have calculated the area of a square. Yes?

 

[Steve4Physics]

edit: Damn! Yes! It's a circle we are talking about, I seem to have calculated the area of a square. Yes?

Yes!

 

[Steve4Physics]

can you please explain the limits of the integral more? I mean, I put a as the limit of the integral as it's a part of a circle we are talking about here, why are you putting a different limit?

Draw ##xy## axes and the 1st quadrant.

Draw a thin vertical ‘elementary’ strip of thickness ##dx## for some value of ##x## inside the quadrant. Note:
- the left side of the strip has x-coordinate ##x##;
- the right side of the strip has x-coordinate ##x+dx##;
- the bottom edge of the strip is a has y-coordinate = 0;
- the top edge of the strip has y-coordinate ##\sqrt {a^2-x^2}## (that’s the key point).

When you integrate ##xy## over the elementary strip you get ##\int_{y=0}^{y=\sqrt {a^2-x^2}} xy~ dy~dx##. (Then it remains to integrate the contributions from all the strips.)

Can you take t from there?

 

[Slimy0233]

Can you take t from there?

It was very kind of you to explain that well and that much!

I can take it from there, thank you! I managed to solve it.