Some questions about double integral and vector calculus

[abcdefg10645]

I've written all the questions in the PDF file ...

Please help me !

 

[HallsofIvy]

If you can't be bothered to
1) type the problem itself here
2) Do anything at all at all on the problem

I don't see why I should bother.

 

[abcdefg10645]

If you can't be bothered to
1) type the problem itself here
2) Do anything at all at all on the problem

I don't see why I should bother.

I'm sorry for that.

I scanned the problem I had just because they are "hard" to type ...I concede that my computer ability is not good... So I made manuscript instead.

And I did not discribed the problem I have ...is my false .

I'm now typing the problem I have...

For Q1:

I tried change of variable and fail...

I could not figure out the craftsmanship solving the problem...

Give me a hint please...

For Q2:

I was considering expand the formula by assuming

A=(A1)i+(A2)j+(A3)k {also B}

substitude the equations...

No doubt ...It is quite terrible !

So I asked for help!

Sorry again for ignoring the "detail problem" :shy:

 

[tiny-tim]

Hi abcdefg10645! That's better!

(have a del: ∇ and a dot: · and an integral: ∫ )

For 1, make the obvious substitution …

u = 3x + y, v = x - 2y, dxdy = … ?

For 2, just write each of the coordinates out in full …

try a. first …

what do you get?

 

[abcdefg10645]

Hi abcdefg10645! That's better!

(have a del: ∇ and a dot: · and an integral: ∫ )

For 1, make the obvious substitution …

u = 3x + y, v = x - 2y, dxdy = … ?

For 2, just write each of the coordinates out in full …

try a. first …

what do you get?

Thaks for your help...

But for question 1...I still have a little problem...

Could you give me a hint for changing the upper and lower limit?

And even tell me the "principle" resoving the limits ...

Thanks!

 

[tiny-tim]

Could you give me a hint for changing the upper and lower limit?

I don't understand …

just convert x and y to u and v …

what do you get?

 

[HallsofIvy]

The region in problem 1 has boundaries 2x+ y= 0, 3x+ y= 0, x- 2y= 1 and x- 2y= 2.

tiny-tim suggested the substitution u = 3x + y, v = x - 2y.

Solve for x and y in terms of u and v: Multiplying the first equation by 2 and adding to the second equation gives 2u+ v= 6x+ 2y+ x- 2y= 7x so x= (2u+ v)/7. Multiplying the second equation by 3 and subtracting the first equation gives 3v- u= 3x-6y- 3x- y= -7y so y= (u- 3v)/7.

Convert the boundary to uv-coordinates by substituting those for x and y:
2x+ y= 2(2u+v)/7+ (u-3v)/7= (4u+ 2v+ u- 3v)/7= (5u- v)/7= 0 so 5u- v= 0

It should be obvious that the last three equations become u= 0, v= 1, and v= 2, but as a demonstration, x- 2y= (2u+ v)/7- 2(u- 3v)/7= (2u+ v- 2u+ 6v)/7= 7v/7= v= 1.

So you want to integrate on the region bounded by the lines u= 0, u= v/5, v= 1, and v= 2. I recommend taking the "outer integral" to be with respect to v, from 1 to 2, and the "inner integral" with respect to u, from 0 to v/5. Be careful about dudv. It is, of course, the Jacobian.

 

[djeitnstine]

try sketching your boundaries