Integration probelm

• To: mathgroup at smc.vnet.net
• Subject: [mg122038] Integration probelm
• From: Jing <jing.guo89 at yahoo.com>
• Date: Mon, 10 Oct 2011 04:27:31 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```Hi,
I am trying to use mathmatica to integrate a pretty complicated expression. But there is some problem for the integration. Seeking help...

My aim is to integrate the following expression:
Integrate[-1/2 y (x + Sqrt[r^2 - y^2]) +
1/4 (-Sqrt[3] x + y) (1/2 (x + Sqrt[3] y) + Sqrt[
r^2 - 1/4 (-Sqrt[3] x + y)^2]) +
1/2 r^2 (-(2 pi)/3 + ArcCos[-y/r] +
ArcCos[(-Sqrt[3] x + y)/(2 r)]), {x, -Sqrt[r^2 - y^2],
y/Sqrt[3] }, {y, -r*Sqrt[3]/2, 0}]

I divide the expression in to 5 parts.

1. Integrate[-1/2 y (x + Sqrt[r^2 - y^2]), {x, -Sqrt[r^2 - y^2],
y/Sqrt[3] },
Assumptions -> {y < 0 && r > 0 && y^2 < r^2},{y, -r*Sqrt[3]/2, 0}, Assumptions -> {y < 0 && r > 0 && y^2 < r^2 &&
r^2 < 4 y^2/3}] // FullSimplify

The answer is -1/144 (-9 + Sqrt[3] \[Pi]) r^4  no problem

2. Integrate[
1/8 (-Sqrt[3] x + y) (x + Sqrt[3] y) , {x, -Sqrt[r^2 - y^2],
y/Sqrt[3] },
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3},{y, -r*Sqrt[3]/2, 0}, Assumptions -> {y < 0 && r > 0 && y^2 < r^2 &&
r^2 < 4 y^2/3}] ]
The answer is -(3 r^4)/64 no problem

3. First I integrate:
Integrate[
1/4 (-Sqrt[3] x + y) Sqrt[
r^2 - 1/4 (-Sqrt[3] x + y)^2], {x, -Sqrt[r^2 - y^2], y/Sqrt[3] },
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 &&
r^2 < 4 y^2/3}] // FullSimplify

output is 1/72 (8 Sqrt[3] r^3 -
Sqrt[3] r^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] -
2 Sqrt[3] y^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] +
6 y Sqrt[(r^2 - y^2) (r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2]))]

Then I integrate :
Integrate[
1/72 (8 Sqrt[3] r^3 -
Sqrt[3] r^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] -
2 Sqrt[3] y^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] +
6 y Sqrt[(r^2 - y^2) (r^2 +
2 y (y - Sqrt[3] Sqrt[r^2 - y^2]))]), {y, -r*Sqrt[3]/2, 0},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]

This time, the Mathmatic can not do integrate, it show me nothing!

4. Integrate[-1/2 r^2 (2 \[Pi] )/3+1/2 r^2*ArcCos[-y/r], {x, -Sqrt[r^2 - y^2], y/Sqrt[3] },
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3},, {y, -r*Sqrt[3]/2, 0}, Assumptions -> {y < 0 && r > 0 && y^2 < r^2 &&
r^2 < 4*y^2/3}]
The answer is -1/72 (-9 + Sqrt[3] \[Pi] + 2 \[Pi]^2) r^4 no probelm.

5. Again, I divide it into two parts.
First I integrate:
1/2 r^2 * ArcCos[(-Sqrt[3] x + y)/(2 r)

( it should alter the variable x to t= (-Sqrt[3] x + y)/(2 r).
in order to simplify the integration :Integrate[
r^3 ArcCos[t]/Sqrt[3], {t, 0, (Sqrt[3 (r^2 - y^2)] + y)/(2 r)},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 &&
r^2 < 4 y^2/3}] // FullSimplify)

Answer is 1/(2 Sqrt[3])r^2 (2 r - Sqrt[
r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] + (y +
Sqrt[3] Sqrt[r^2 - y^2]) ArcSec[(2 r)/(
y + Sqrt[3] Sqrt[r^2 - y^2])])

Then, problem occurs again :
Integrate[
1/(2 Sqrt[3])
r^2 (2 r - Sqrt[
r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] + (y +
Sqrt[3] Sqrt[r^2 - y^2]) ArcSec[(2 r)/(
y + Sqrt[3] Sqrt[r^2 - y^2])]), {y, -r*Sqrt[3]/2, 0},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]

It can not work. and no answer.

Based on what I have done, I find that the Mathmatic can not integrate the following expression:

1. Integrate[
1/(2 Sqrt[3])
r^2 (-Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])]), {y, -r*
Sqrt[3]/2, 0},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]

2. Integrate[
(y + Sqrt[3] Sqrt[r^2 - y^2]) ArcSec[(2 r)/(
y + Sqrt[3] Sqrt[r^2 - y^2])], {y, -r*Sqrt[3]/2, 0},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]

3.Integrate[
1/72 (8 Sqrt[3] r^3 -
Sqrt[3] r^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] -
2 Sqrt[3] y^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] +
6 y Sqrt[(r^2 - y^2) (r^2 +
2 y (y - Sqrt[3] Sqrt[r^2 - y^2]))]), {y, -r*Sqrt[3]/2, 0},
Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]

When I try to integrate the above equations, the Mathmatic doesn't work.

I hope someone can help me to figure it out.

Cheers

```

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