Problem 19 - Calculus an Applied Approach

Question

Evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{\sqrt{1-y^{2}}}-5 x y d x d y $$

Step-by-Step Solution

Answer

The value of the given double integral is -10.

Step 1: Perform Inner Integration

Firstly, compute the integral with respect to \(x\) in the given interval, which is from 0 to \(\sqrt{1-y^2}\):\n\[\int_{0}^{\sqrt{1-y^{2}}}-5yx \, dx = [-2.5xy^2]_{0}^{\sqrt{1-y^2}} = -2.5y(\sqrt{1-y^2})^2 + 0 = -2.5y(1-y^2) = -2.5y + 2.5y^3\].

Step 2: Perform Outer Integration

Secondly, perform the integral with respect to \(y\) in the interval from 0 to 2:\n\[\int_{0}^{2} (-2.5y + 2.5y^3) dy = [-1.25y^2 + 0.625y^4]_{0}^{2} = -1.25(2)^2 + 0.625(2)^4 = -10\].

Step 3: Conclusion

As a result, after performing both integrations carefully and correctly, the double integral of the given function is found to be equal to -10.

Key Concepts

Inner Integration

In double integrals, the first step usually involves evaluating the inner integral. This is like working with a one-dimensional integral, but within a specified range for another variable. Here, we deal with the variable under the innermost integral sign. It's crucial to focus on the order of integration, as it affects the integral's bounds and sometimes the method of integration.
For instance, we need to perform the inner integration of the function \(-5xy\) with respect to \(x\), from \(0\) to \(\sqrt{1-y^2}\). This step simplifies our problem, as it reduces the function's dimensionality before proceeding to the next integration. Calculating this involves substituting the limits back into the integrated function.
So, we evaluate the integral:

• Integrate \(-5xy\) with respect to \(x\).
• This becomes \(-2.5xy^2\) evaluated from \(0\) to \(\sqrt{1-y^2}\).
• After substitution, we simplify it to \(-2.5y(1-y^2) = -2.5y + 2.5y^3\).

This step reduces the complexity by focusing on a single variable, preparing the function for the outer integration.

Outer Integration

After the inner integration, we proceed with the outer integration. This involves integrating the result from the inner step with respect to the second variable, typically the one originally outside the integral signs. This step finalizes the double integration process and determines the final value of the integral.
In our specific example, we integrate \(-2.5y + 2.5y^3\) with respect to \(y\), over the interval from 0 to 2. The outer integration involves substituting these bounds after integrating the function. Remember to follow the order of operations, as cumbersome as it might seem, it ensures a correct solution.
Steps to follow for outer integration:

• Integrate the function \(-2.5y + 2.5y^3\) with respect to \(y\).
• This evaluates to \([-1.25y^2 + 0.625y^4]\) at the limits from 0 to 2.
• Substituting the limits gives us the final answer: \(-10\).

The outer integration reconciles both variables in the double integral, delivering the overall solution to the given problem.

Integral with respect to x

Integrating with respect to \(x\) means treating \(x\) as the variable and keeping others as constants. This is typical for the inner integration step when it comes to double integrals. The process makes use of the bounds specified for \(x\), which are crucial for correctly evaluating the integral.
When we compute the integral \(-5yx\) with respect to \(x\) over the interval from 0 to \(\sqrt{1-y^2}\), we follow these steps:

• The variable \(y\) is treated as a constant as it does not change within this interval.
• Integrate \(-5xy\) with respect to \(x\). This results in \(-2.5xy^2\).
• Substitute the limits of \(x\) to refine the result.

This approach simplifies the challenge of handling two variables at once by initially isolating one, allowing us to focus our computations.

Integral with respect to y

Integrating with respect to \(y\) requires treating \(y\) as the variable and considering all other terms as constants. This approach is typically associated with the outer integration process during which all earlier computations start forming the final solution. The precision in bounds application for \(y\) significantly impacts the accuracy of the integration outcome.
For our problem, this means integrating \(-2.5y + 2.5y^3\) with respect to \(y\) from 0 to 2. Here's a basic rundown of what happens:

• Shift perspective to make \(y\) the focal variable.
• Compute the integral, resulting in \([-1.25y^2 + 0.625y^4]\).
• Apply the specified interval limits (0 to 2) to yield the final value: \(-10\).

This integral essentially wraps up the composite areas defined by the double integral, cementing the entire process into a single numerical answer.