Trigonometric identity double definite integral

• clairaut
In summary: To parametrise the rectangle, you need two functions x(u, v) and y(u, v) where u and v are parameters. You have to choose the parameters so that they cover the rectangle 2\le x\le 4, 2\le y\le 6. For example, you can choose u = x and v = y. This way, u covers the interval [2, 4] and v covers the interval [2, 6]. So you can usex = u, y = v.You can then make the substitution u = r cosθ and v = r sinθ to get a double integral in polar coordinates. However, this may not be
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.

clairaut said:
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.

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

ehild

Where am I supposed to start?

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

clairaut said:
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.

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.

Thanks. I'll try it. It looks nasty

I've tried

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

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

I can't get the substitution to match

I like what pasmith said.

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

Also don't forget to change your limits.

clairaut said:
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.

1. What is a trigonometric identity?

A trigonometric identity is an equation that is true for all values of the variables involved. In other words, it is an equation that holds true regardless of the specific values of the angles or other variables involved.

2. What is a double definite integral?

A double definite integral is an integral that involves two variables and has both upper and lower limits of integration. This means that the integral is calculated over a specific region in the two-dimensional plane.

3. How are trigonometric identities used in double definite integrals?

Trigonometric identities can be used to simplify the integrand in a double definite integral. This is especially useful when the integral involves trigonometric functions and can make the integration process more manageable.

4. What are some common trigonometric identities used in double definite integrals?

Some common trigonometric identities used in double definite integrals include the double angle identities, half angle identities, and the Pythagorean identities. These identities can be used to simplify the integrand and make the integration process easier.

5. How do you solve a double definite integral using trigonometric identities?

To solve a double definite integral using trigonometric identities, first simplify the integrand using the identities. Then, use the appropriate techniques for integration, such as u-substitution or integration by parts, to solve the integral. Finally, evaluate the integral using the given limits of integration to obtain the final answer.

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