- #1

- 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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter clairaut
- Start date

- #1

- 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

- 1,912

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

- 0

- #5

pasmith

Homework Helper

- 2,019

- 651

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

- 0

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

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

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

I can't get the substitution to match

- #8

BiGyElLoWhAt

Gold Member

- 1,573

- 118

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

- 1,573

- 118

Also don't forget to change your limits.

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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

Share: