Trigonometric identity double definite integral

[clairaut]

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.

 

[ehild]

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

 

[clairaut]

Where am I supposed to start?

 

[clairaut]

And why can't I assume the trig substitutions here? Is it because the substitution results in that zero?

 

[pasmith]

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 $$\int_2^4 \left( \int_2^6 \sqrt{52 - x^2 - y^2}\,dy\right)\,dx.$$

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

However I suspect that the $x$-integral you obtain will be nasty.

You can instead set $x = r \cos \theta$, $y = r\sin\theta$, but the limits of the $r$-integral depend on $\theta$ in a complicated way.

 

[clairaut]

Thanks. I'll try it. It looks nasty

 

[clairaut]

I've tried

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

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

I can't get the substitution to match

 

[BiGyElLoWhAt]

I like what pasmith said.

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

 

[BiGyElLoWhAt]

Also don't forget to change your limits.

 

[HallsofIvy]

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 $2\le x\le 4$, $2\le y\le 6$. They do NOT lie on a circle of radius 52 about the origin.