- #1

clairaut

- 72

- 0

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.

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- Thread starter clairaut
- Start date

- #1

clairaut

- 72

- 0

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.

- #2

ehild

Homework Helper

- 15,543

- 1,915

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

clairaut

- 72

- 0

Where am I supposed to start?

- #4

clairaut

- 72

- 0

- #5

pasmith

Homework Helper

- 2,338

- 942

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

clairaut

- 72

- 0

Thanks. I'll try it. It looks nasty

- #7

clairaut

- 72

- 0

I've tried

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

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

I can't get the substitution to match

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

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

I can't get the substitution to match

- #8

BiGyElLoWhAt

Gold Member

- 1,577

- 119

I like what pasmith said.

Via pythagoreans theorem -(x^2 + y^2) = -r^2

Via pythagoreans theorem -(x^2 + y^2) = -r^2

- #9

BiGyElLoWhAt

Gold Member

- 1,577

- 119

Also don't forget to change your limits.

- #10

HallsofIvy

Science Advisor

Homework Helper

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- 970

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.I've tried

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

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

I can't get the substitution to match

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