Problem 16 - University Calculus: Early Transcendentals

Question

Evaluate the double integral over the given region \(R\). $$\iint_{R}\left(\frac{\sqrt{x}}{y^{2}}\right) d A, \quad R: \quad 0 \leq x \leq 4, \quad 1 \leq y \leq 2$$

Step-by-Step Solution

Answer

The value of the double integral is \(\frac{8}{3}\).

Step 1: Set up the Double Integral

The given region is described by the inequalities \(0 \leq x \leq 4\) and \(1 \leq y \leq 2\). Therefore, the double integral can be set up as: \[ \int_{1}^{2} \int_{0}^{4} \frac{\sqrt{x}}{y^2} \, dx \, dy. \]

Step 2: Integrate with respect to x

First, solve the inner integral \(\int_{0}^{4} \frac{\sqrt{x}}{y^2} \, dx\). Since \(\frac{1}{y^2}\) is a constant with respect to \(x\), it can be factored out of the integral as follows: \(\frac{1}{y^2} \int_{0}^{4} \sqrt{x} \, dx\). The integral \(\int_{0}^{4} \sqrt{x} \, dx\) can be evaluated as: \[\int_{0}^{4} x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = \frac{2}{3} \times 8 = \frac{16}{3}.\]

Step 3: Substitute and Simplify

Substitute the result from Step 2 into the original integral: \(\int_{1}^{2} \frac{16/3}{y^2} \, dy\). Simplify it further to \(\frac{16}{3} \int_{1}^{2} y^{-2} \, dy\).

Step 4: Integrate with respect to y

Evaluate the integral \(\int_{1}^{2} y^{-2} \, dy\) by finding the antiderivative of \(y^{-2}\), which is \(-y^{-1}\). Thus, \[\left[ -y^{-1} \right]_{1}^{2} = -\frac{1}{2} + 1 = \frac{1}{2}.\]

Step 5: Final Calculation

Multiply the result of the \(y\)-integral by the constant factor obtained after the \(x\)-integration. Thus, \(\frac{16}{3} \times \frac{1}{2} = \frac{16}{6} = \frac{8}{3}.\)

Key Concepts

Integration with respect to x

When evaluating a double integral, the typical process begins with integrating with respect to one variable at a time. In our exercise, we first integrate with respect to the variable \(x\). This means we consider \(y\) as a constant during this part of the integration.
This approach simplifies our integrand, allowing us to focus only on the part of the function that contains \(x\).

• The double integral we are dealing with is \(\int_{1}^{2} \int_{0}^{4} \frac{\sqrt{x}}{y^2} \, dx \, dy\).
• We start by focusing on the inner integral \(\int_{0}^{4} \frac{\sqrt{x}}{y^2} \, dx\).
• Here, \(\frac{1}{y^2}\) is a constant because it does not depend on \(x\), allowing us to treat it as such.
• This simplifies to \(\frac{1}{y^2} \int_{0}^{4} \sqrt{x} \, dx\).
• You compute the integral, resulting in: \[ \int_{0}^{4} x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = \frac{16}{3}. \]

By calculating this integral, we find a more simplified expression to be used in the further steps of our computation.

Integration with respect to y

The next step involves integrating with respect to \(y\). After performing the \(x\)-integration, we substitute the result into the outer integral. This process requires us to find the antiderivative with respect to \(y\).
Here's how it goes:

• After the \(x\)-integration, we place the result, \(\frac{16}{3}\), back into the integral for \(y\): \(\int_{1}^{2} \frac{16/3}{y^2} \, dy\).
• The integral now simplifies to: \(\frac{16}{3} \int_{1}^{2} y^{-2} \, dy\).
• We need the antiderivative of \(y^{-2}\), which is \(-y^{-1}\).
• Evaluating the integral gives: \[ \left[ -y^{-1} \right]_{1}^{2} = -\frac{1}{2} + 1 = \frac{1}{2}. \]

This outcome shows us the cumulative effect the range of \(y\) has when applied to our previously calculated result.

Antiderivative

The concept of an antiderivative is foundational in calculus, especially when dealing with integration. An antiderivative of a function is another function whose derivative returns the original function.
Being able to find the antiderivative is crucial for solving integrals. In the context of our double integral, finding antiderivatives is essential for both the \(x\) and \(y\) integrations.

• For \(x\), the antiderivative of \(x^{1/2}\) is \(\frac{2}{3}x^{3/2}\), as we see when we compute \(\int_{0}^{4} x^{1/2} \, dx\).
• For \(y\), the antiderivative of \(y^{-2}\) is \(-y^{-1}\), highlighted in our evaluation of \(\int_{1}^{2} y^{-2} \, dy\).

Without finding these antiderivatives, evaluating the definite integrals at their limits would be impossible. The process links tangible computation back to the theory of differentiation, reinforcing how integral and derivative concepts are interconnected in mathematics.