Double integral over triangular region

[anniecvc]

Homework Statement

Integrate f(u,v)= v - sqrt(u) over the triangular region cut from the first quadrant by the line u+v=64 in the uv plane.

Homework Equations

I am assuming u is the equivalent of the x-axis in the xy plane and v the equivalent of y in the xy plane.
I am taking the triangle as a Type I region.

The Attempt at a Solution

limits of integration:

0≤u≤65, 0≤v≤64-u

∫ ∫ v-sqrt(u) dvdu

∫ (1/2)v² - sqrt(u)v evaluated from v=0 to v=64-u

∫ 2048 - 64u + (1/2)u² - 64*u^1/2 + u^3/2

2048u - 32u² + (1/6)u³ - (2/3)*64*u^3/2+(2/5)*u^5/2 evaluated from u=0 to u=64

I get a really narley fraction, namely, 4638576/30, which is of course wrong.

 

[Jufro]

Everything looks good, I double checked your calculations and came to a different result. Double check the final evaluation. I came up with ~ 8,000 (the exact value is for you to figure out).

 

[haruspex]

Everything looks good, I double checked your calculations and came to a different result. Double check the final evaluation. I came up with ~ 8,000 (the exact value is for you to figure out).

I get more like 32000.

 

[anniecvc]

Yes, the exact answer is 524288/15 ~ 35000.

This was a terrible problem not because it was difficult but because the numbers were so ugly.

 

[haruspex]

Yes, the exact answer is 524288/15 ~ 35000.

This was a terrible problem not because it was difficult but because the numbers were so ugly.

It's not so bad if you leave as much as possible in powers of 2.
2¹¹u - 2⁵u² + (1/6)u³ - (2/3)*2⁶*u^3/2+(2/5)*u^5/2 where u = 2⁶:
2¹⁷ - 2¹⁷ + (1/6)2¹⁸ - (2/3)*2⁶*2⁹+(2/5)*2¹⁵ = 2¹⁶{(2/3) - (1/3)+(1/5)} = 2¹⁹/15