# Trigonometric identity double definite integral

1. Jul 8, 2014

### 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.

2. Jul 8, 2014

### ehild

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

ehild

3. Jul 8, 2014

### clairaut

Where am I supposed to start?

4. Jul 8, 2014

### clairaut

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

5. Jul 8, 2014

### pasmith

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.

6. Jul 8, 2014

### clairaut

Thanks. I'll try it. It looks nasty

7. Jul 9, 2014

### clairaut

I've tried

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

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

I can't get the substitution to match

8. Jul 9, 2014

### BiGyElLoWhAt

I like what pasmith said.

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

9. Jul 9, 2014

### BiGyElLoWhAt

Also don't forget to change your limits.

10. Jul 9, 2014

### HallsofIvy

Staff Emeritus
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.