Trigonometric identity double definite integral

  • Thread starter clairaut
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  • #1
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Double integral of (52-x^2-y^2)^.5

2<_ x <_ 4
2<_ y <_ 6

I get up to this simplicity that results in a zero!!!

1-cos^2(@) - sin^2(@) = 0

This identity seems to be useless.

HELP PLEASE.
 

Answers and Replies

  • #2
ehild
Homework Helper
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Double integral of (52-x^2-y^2)^.5

2<_ x <_ 4
2<_ y <_ 6

I get up to this simplicity that results in a zero!!!

1-cos^2(@) - sin^2(@) = 0

This identity seems to be useless.

HELP PLEASE.

The identity is true for angles. But you can not take x=cosθ and y=sinθ here.

ehild
 
  • #3
72
0
Where am I supposed to start?
 
  • #4
72
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And why can't I assume the trig substitutions here? Is it because the substitution results in that zero?
 
  • #5
pasmith
Homework Helper
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Double integral of (52-x^2-y^2)^.5

2<_ x <_ 4
2<_ y <_ 6

I get up to this simplicity that results in a zero!!!

1-cos^2(@) - sin^2(@) = 0

This identity seems to be useless.

HELP PLEASE.

Your integral is [tex]
\int_2^4 \left( \int_2^6 \sqrt{52 - x^2 - y^2}\,dy\right)\,dx.
[/tex]

To do the inner integral: substitute [itex]y = \sqrt{52 - x^2}\sin\theta[/itex]. Remember that for this integral [itex]x[/itex] is a constant.

However I suspect that the [itex]x[/itex]-integral you obtain will be nasty.

You can instead set [itex]x = r \cos \theta[/itex], [itex]y = r\sin\theta[/itex], but the limits of the [itex]r[/itex]-integral depend on [itex]\theta[/itex] in a complicated way.
 
  • #6
72
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Thanks. I'll try it. It looks nasty
 
  • #7
72
0
I've tried

X=(52)^.5 [cos(@)]

Y=(52)^.5 [sin(@)]

I can't get the substitution to match
 
  • #8
BiGyElLoWhAt
Gold Member
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118
I like what pasmith said.

Via pythagoreans theorem -(x^2 + y^2) = -r^2
 
  • #9
BiGyElLoWhAt
Gold Member
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118
Also don't forget to change your limits.
 
  • #10
HallsofIvy
Science Advisor
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I've tried

X=(52)^.5 [cos(@)]

Y=(52)^.5 [sin(@)]

I can't get the substitution to match
x and y are independent variables. (x, y) can be any point on the rectangle [itex]2\le x\le 4[/itex], [itex]2\le y\le 6[/itex]. They do NOT lie on a circle of radius 52 about the origin.
 

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