D={(x,y)|0=<x=<π^2, √(x)<=y<=π}として、∫∫D cos(x/y)dxdy

Given the region \( D = \{(x, y) \mid 0 \leq x \leq \pi^2, \sqrt{x} \leq y \leq \pi \} \), we need to evaluate the double integral:

\[
\iint_D \cos\left(\frac{x}{y}\right) \, dx \, dy
\]

We can express this integral as:

\[
\int_{y=\sqrt{x}}^{\pi} \int_{x=0}^{y^2} \cos\left(\frac{x}{y}\right) \, dx \, dy
\]

First, we evaluate the inner integral with respect to \( x \):

\[
\int_0^{y^2} \cos\left(\frac{x}{y}\right) \, dx
\]

Let \( u = \frac{x}{y} \). Then \( du = \frac{dx}{y} \) or \( dx = y \, du \). The limits of integration for \( u \) are from \( 0 \) to \( y \) (when \( x \) goes from \( 0 \) to \( y^2 \)):

\[
\int_0^{y^2} \cos\left(\frac{x}{y}\right) \, dx = \int_0^y \cos(u) \cdot y \, du = y \int_0^y \cos(u) \, du
\]

The integral of \( \cos(u) \) is \( \sin(u) \):

\[
y \int_0^y \cos(u) \, du = y \left[ \sin(u) \right]_0^y = y \left( \sin(y) - \sin(0) \right) = y \sin(y)
\]

So the inner integral evaluates to:

\[
\int_0^{y^2} \cos\left(\frac{x}{y}\right) \, dx = y \sin(y)
\]

Next, we evaluate the outer integral with respect to \( y \):

\[
\int_{\sqrt{x}}^{\pi} y \sin(y) \, dy
\]

The limits of integration for \( y \) are from \( 0 \) to \( \pi \):

\[
\int_0^{\pi} y \sin(y) \, dy
\]

We use integration by parts to solve this integral. Let \( u = y \) and \( dv = \sin(y) \, dy \). Then \( du = dy \) and \( v = -\cos(y) \):

\[
\int y \sin(y) \, dy = -y \cos(y) \bigg|_0^{\pi} + \int_0^{\pi} \cos(y) \, dy
\]

Evaluating the boundary term:

\[
-y \cos(y) \bigg|_0^{\pi} = -\pi \cos(\pi) + 0 \cdot \cos(0) = -\pi (-1) + 0 = \pi
\]

Now, evaluate the integral of \( \cos(y) \):

\[
\int_0^{\pi} \cos(y) \, dy = \sin(y) \bigg|_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0
\]

Adding the two parts together:

\[
\int_0^{\pi} y \sin(y) \, dy = \pi + 0 = \pi
\]

Thus, the value of the double integral is:

\[
\iint_D \cos\left(\frac{x}{y}\right) \, dx \, dy = \pi
\]

Would you like any details or have any questions?

Here are 8 related questions to expand on this topic:
1. How does the region \( D \) affect the limits of integration?
2. Can you explain the change of variables technique used in solving the integral?
3. What are some common methods for evaluating double integrals?
4. How does integration by parts work for definite integrals?
5. What are the applications of double integrals in physics?
6. How do you interpret the result of this double integral geometrically?
7. Can you solve a similar problem with a different integrand, like \( \sin\left(\frac{x}{y}\right) \)?
8. How do the properties of trigonometric functions influence the integration process?

**Tip:** When solving double integrals, sketching the region of integration can provide valuable insights into setting up the correct limits of integration.

Learn how to evaluate the double integral ∫∫D cos(x/y) dxdy over the region D={(x,y)|0≤x≤π^2, √(x)≤y≤π}. This detailed solution covers the use of change of variables and integration by parts techniques. Suitable for advanced college-level students and mathematics enthusiasts.

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