To solve these problems, we must reverse the order of integration for each given double integral. Let’s tackle them one by one.

Problem (1)

$\int_0^1 \int_{x}^1 \sqrt{y} \sin(y) \, dy \, dx.$

Step 1: Understand the current limits

The integration order is $dy$ first and $dx$ second. For $y$, the limits are $x \leq y \leq 1$, and for $x$, the limits are $0 \leq x \leq 1$.

Step 2: Reverse the order of integration

To reverse, we determine the bounds:

• For $x$, the region is described by $0 \leq x \leq y$ (as $x \leq y$ from the original inner limit).
• For $y$, the region is $0 \leq y \leq 1$.

Rewriting the integral: $\int_0^1 \int_0^y \sqrt{y} \sin(y) \, dx \, dy.$

Step 3: Evaluate the inner integral

The inner integral with respect to $x$ is: $\int_0^y \sqrt{y} \sin(y) \, dx = \sqrt{y} \sin(y) \cdot \int_0^y 1 \, dx = \sqrt{y} \sin(y) \cdot y.$

So the integral becomes: $\int_0^1 y \sqrt{y} \sin(y) \, dy.$

Simplify $y \sqrt{y} = y^{3/2}$, so: $\int_0^1 y^{3/2} \sin(y) \, dy.$

This integral can now be solved using integration by parts or numerical methods if needed.

Problem (2)

$\int_0^1 \int_{\sqrt{x}}^1 \sqrt{y^3 + 1} \, dy \, dx.$

Step 1: Understand the current limits

For $y$, the limits are $\sqrt{x} \leq y \leq 1$, and for $x$, the limits are $0 \leq x \leq 1$.

Step 2: Reverse the order of integration

• For $x$, since $y \geq \sqrt{x}$, we have $0 \leq x \leq y^2$.
• For $y$, the region is $0 \leq y \leq 1$.

Rewriting the integral: $\int_0^1 \int_0^{y^2} \sqrt{y^3 + 1} \, dx \, dy.$

Step 3: Evaluate the inner integral

The inner integral with respect to $x$ is: $\int_0^{y^2} \sqrt{y^3 + 1} \, dx = \sqrt{y^3 + 1} \cdot \int_0^{y^2} 1 \, dx = \sqrt{y^3 + 1} \cdot y^2.$

The integral becomes: $\int_0^1 y^2 \sqrt{y^3 + 1} \, dy.$

This integral can be solved using substitution methods.

Problem (3)

$\int_0^8 \int_{2\sqrt{x/\pi}}^2 e^{x^4} \, dx \, dy.$

Step 1: Understand the current limits

For $x$, the limits are $2\sqrt{\frac{x}{\pi}} \leq y \leq 2$, and for $y$, the limits are $0 \leq x \leq 8$.

Step 2: Reverse the order of integration

For $y$, since $y \leq 2$ and $y \geq 2\sqrt{\frac{x}{\pi}}$, rearranging gives $x \leq \frac{\pi y^2}{4}$.

So:

• For $x$, the region is $0 \leq x \leq 8$.
• For $y$, the region is $0 \leq y \leq 2$.

Rewriting the integral: $\int_0^2 \int_0^{\pi y^2/4} e^{x^4} \, dx \, dy.$

Step 3: Evaluate the inner integral

The inner integral with respect to $x$ is not elementary: $\int_0^{\pi y^2/4} e^{x^4} \, dx.$

This can be evaluated numerically, but the limits have been successfully reversed.

Would you like detailed steps for solving the integrals further? Let me know!

Related Questions

1. How do you find the region of integration for a double integral?
2. What are the advantages of reversing the order of integration?
3. How do you solve integrals involving non-elementary functions?
4. What techniques can simplify integration after reversing the limits?
5. What role does symmetry play in reversing integration limits?

Tip

When reversing the order of integration, always sketch the region to ensure the limits are correctly swapped.