To evaluate the given integral

$\iint_D x \cos y \, dA$

where $D$ is the region bounded by $y = 0$, $y = x^2$, and $x = 1$, we proceed as follows:

Step 1: Analyze the region $D$

• $y = 0$ is the lower boundary.
• $y = x^2$ is the parabolic upper boundary.
• $x = 1$ is the vertical line that bounds the region on the right.

The region $D$ is therefore defined as: $0 \leq x \leq 1, \quad 0 \leq y \leq x^2.$

Step 2: Set up the integral

Using the bounds for $D$, the integral is written as: $\int_0^1 \int_0^{x^2} x \cos y \, dy \, dx.$

Step 3: Evaluate the inner integral

The inner integral is: $\int_0^{x^2} x \cos y \, dy.$

Since $x$ is constant with respect to $y$, factor $x$ out: $x \int_0^{x^2} \cos y \, dy.$

The integral of $\cos y$ is $\sin y$, so: $x \left[\sin y \right]_0^{x^2} = x \left(\sin(x^2) - \sin(0)\right).$

Since $\sin(0) = 0$, this simplifies to: $x \sin(x^2).$

Step 4: Evaluate the outer integral

The outer integral is: $\int_0^1 x \sin(x^2) \, dx.$

Let $u = x^2$, so $du = 2x \, dx$, or $\frac{du}{2} = x \, dx$.

When $x = 0$, $u = 0$. When $x = 1$, $u = 1$.

The integral becomes: $\int_0^1 x \sin(x^2) \, dx = \frac{1}{2} \int_0^1 \sin u \, du.$

The integral of $\sin u$ is $-\cos u$, so: [ \frac{1}{2} \left[-\cos u \right]_0^1 = \frac{1}{2} \left[-\cos(1) + \cos(0)\right]. ]

Since $\cos(0) = 1$, this simplifies to: $\frac{1}{2} \left[-\cos(1) + 1\right] = \frac{1 - \cos(1)}{2}.$

Final Answer:

$\frac{1 - \cos(1)}{2}.$

Would you like further details or clarification?

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Related Questions:

1. How do you evaluate similar double integrals over non-rectangular regions?
2. What substitution techniques simplify integrals involving trigonometric functions?
3. How do you visualize the region of integration for double integrals?
4. Can this integral be solved using polar coordinates?
5. What are the applications of such integrals in physics and engineering?

Tip:

For integrals involving $x \sin(x^2)$ or similar terms, substitution is often the most efficient method.