Expert: Paul Klarreich - 12/13/2006

Question
can you help with these 5 questions these are final questions thank you.

http://img138.imageshack.us/img138/2040/untitled3pz9.jpg
http://img138.imageshack.us/img138/9012/untitled4jy3.jpg


Answer
Questioner:  Paul
Category:  Calculus
 
Subject:  calculus help
Question:  can you help with these 5 questions these are final questions thank you.

http://img138.imageshack.us/img138/2040/untitled3pz9.jpg
http://img138.imageshack.us/img138/9012/untitled4jy3.jpg

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An element of polar area is given by the formula for area of a sector.

dA = 1/2 r^2 d@

In this example,  r = 2 sin @ + 4 cos @
and you integrate for  @ (theta) = 0 to pi/2
{
| 1/2(2 sin @ + 4 cos @ )^2 d@
}

{
| 1/2(4 sin^2 @ + 16 cos^2 @ + 16 sin @ cos @ ) d@
}

{
| (2 sin^2 @ + 8 cos^2 @ + 8 sin @ cos @ ) d@
}

Half-angle formulas used here. (Double-angle, if you like.)

{
| (1 - cos 2@ + 4(1 + cos 2@) + 4 sin 2@ ) d@
}

{
| (1 - cos 2@ + 4 + 4 cos 2@ + 4 sin 2@ ) d@
}

{
| (5 + 3 cos 2@ + 4 sin 2@ ) d@
}

5@ + 3/2 sin 2@ - 2 cos 2@,  from  0 to pi/2

(5pi/2 + 0 + 2) - (0 + 0 - 2(-1))

5pi/2 + 2 - 2

5pi/2
.......................................

The arc length in P.C. is:

{
| sqrt( r^2 + (dr/d@)^2) d@
}

Now dr/d@ = 2 cos @ - 4 sin @

r^2        = 4 sin^2 @ + 16 cos^2 @ + 16 sin @ cos @  from above.
(dr/d@)^2  = 4 cos^2 @ + 16 sin^2 @ - 16 sin @ cos @  from above.

And their sum is:

4(sin^2 @ + cos^2 @) + 16( same ) =  20.

Integrate:

{
| sqrt(20) d@
}

{
| 2 sqrt(5) d@
}


= 2 sqrt(5) @,  from  0 to pi/2

= pi sqrt(5)
....................................

I THINK THAT'S ENOUGH.