Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{0}^{9} \int_{y / 3}^{\sqrt{y}}\left(x y^{2}\right) d x d y $$

Short answer

(a) The value of the iterated integral is 164.025. (b) The integral rewritten with the reverse order of integration is \( \int_{0}^{3} \int_{3x}^{x^2} xy^2 \, dy \, dx \).

Step-by-step solution

Evaluate the Inner Integral

First, let's evaluate the inner integral with respect to \( x \): \[\int_{y/3}^{\sqrt{y}} x y^2 \, dx\]Since \( y^2 \) is treated as a constant with respect to \( x \), the integral simplifies to:\[y^2 \int_{y/3}^{\sqrt{y}} x \, dx\]The antiderivative of \( x \) is \( \frac{x^2}{2} \).Thus, the integral becomes:\[y^2 \cdot \left[ \frac{x^2}{2} \right]_{y/3}^{\sqrt{y}} = y^2 \cdot \left( \frac{(\sqrt{y})^2}{2} - \frac{(y/3)^2}{2} \right)\]Evaluating this gives:\[y^2 \cdot \left( \frac{y}{2} - \frac{y^2/9}{2} \right)= y^2 \cdot \left( \frac{9y - y^2}{18} \right)\]This simplifies to:\[\frac{y^2 (9y - y^2)}{18}\]

Evaluate the Outer Integral

Now, substituting the result of the inner integral into the outer integral:\[\int_{0}^{9} \frac{y^2 (9y - y^2)}{18} \, dy\]Distribute \( y^2 \):\[\int_{0}^{9} \frac{9y^3 - y^4}{18} \, dy\]Simplify to:\[\frac{1}{18} \int_{0}^{9} (9y^3 - y^4) \, dy\]Break into separate integrals:\[\frac{1}{18} \left( 9 \int_{0}^{9} y^3 \, dy - \int_{0}^{9} y^4 \, dy \right)\]Solve each integral:\[\int y^3 \, dy = \frac{y^4}{4}; \int y^4 \, dy = \frac{y^5}{5}\]Substituting back, we have:\[\frac{1}{18} \left( 9 \cdot \left[ \frac{y^4}{4} \right]_{0}^{9} - \left[ \frac{y^5}{5} \right]_{0}^{9} \right)\]Compute the bounds:\[\frac{1}{18} \left( 9 \cdot \frac{9^4}{4} - \frac{9^5}{5} \right)\]Calculate values:\[9^4 = 6561; \quad 9^5 = 59049\]Substitute and compute:\[\frac{1}{18} \left( 9 \cdot \frac{6561}{4} - \frac{59049}{5} \right) = \frac{1}{18} \left( 14762.25 - 11809.8 \right)\]The result is:\[\frac{1}{18} \times 2952.45 = 164.025\]

Rewrite the Integral Using the Other Order of Integration

Determine the region of integration:- Boundaries of \( y \) are from 0 to 9.- For a given \( y \), \( x \) ranges from \( y/3 \) to \( \sqrt{y} \).Consider reversing the integration order by finding the new bounds for \( x \) and \( y \). The intersection points \( y/3 = \sqrt{y} \) determine the critical points for \( y \):\[y = 0 \quad \text{or} \quad y = 9\]Re-evaluate for \( x \) bounds:- For each \( x \), find where \( y/3 < x < \sqrt{y} \) is true, thus for \( x^2 = y \), change it to the corresponding \( y \) as \( 3x < y < x^2 \). So the double integral with changed limits becomes:\[\int_{0}^{3} \int_{3x}^{x^2} xy^2 \, dy \, dx\]

Key concepts

Order of Integration

The order of integration refers to the sequence in which you perform the integrals in an iterated integral. In iterated integrals, you typically have multiple integrals that need to be solved successively. The given iterated integral starts with integration over \( x \) first, followed by \( y \), denoted as \( \int_{0}^{9} \int_{y / 3}^{\sqrt{y}} (x y^2) \, dx \, dy \). Here, the order of integration is "dx dy," which means you evaluate the inner integral with respect to \( x \) first and then tackle the outer integral with respect to \( y \).

If you reverse the order of integration, you solve the integral with respect to \( y \) first and then \( x \). This requires redefining the limits of integration, as illustrated in the solution when the integral is rewritten as \( \int_{0}^{3} \int_{3x}^{x^2} xy^2 \, dy \, dx \). Changing the order of integration can simplify calculations and clarify which parts of the function are independent variables during integration.

Antiderivative

Finding the antiderivative is a crucial step in solving an integral. The antiderivative is the opposite of differentiation. It involves finding a function whose derivative matches the integrand. In the solution given, when evaluating the inner integral \( \int_{y/3}^{\sqrt{y}} x \, dx \), the antiderivative of \( x \) is calculated as \( \frac{x^2}{2} \).

Recognizing the antiderivative allows you to apply the Fundamental Theorem of Calculus, which simplifies the integration process and helps in evaluating definite integrals. The process involves substituting the upper and lower bounds of the variable of integration into the antiderivative, then computing the difference.

Inner Integral

The inner integral refers to the integral that you solve first in an iterated integral. In the expression \( \int_{0}^{9} \int_{y / 3}^{\sqrt{y}} (x y^2) \, dx \, dy \), the inner integral \( \int_{y/3}^{\sqrt{y}} x y^2 \, dx \) needs to be solved before progressing to the outer integral.

In the solution process, the term \( y^2 \) is treated as a constant during the evaluation of the inner integral. This simplification allows you to focus on the variable \( x \) alone, turning the integral into a simpler form \( y^2 \int_{y/3}^{\sqrt{y}} x \, dx \).

Solving the inner integral first is crucial before proceeding to the outer integral as it directly influences the limits and values used in subsequent calculations.

Outer Integral

Once the inner integral is computed, the next step is the outer integral. This integral uses the result of the inner integral and applies it to the remaining variables. Referring to our original expression \( \int_{0}^{9} \left( y^2 \cdot \left( \frac{y}{2} - \frac{y^2/9}{2} \right) \right) \ dy \), we simplify it to \( \int_{0}^{9} \frac{y^2 (9y - y^2)}{18} \, dy \).

The outer integral aggregates the results across the entire range specified for the outer variable, which is \( y \) in this scenario. This step involves breaking down multifaceted expressions into separate integrals that are more manageable, as seen in \( \frac{1}{18} \left( 9 \int_{0}^{9} y^3 \, dy - \int_{0}^{9} y^4 \, dy \right) \), simplifying integration through separate evaluations.

Completing the outer integral gives a final value for the iterated integral, providing a solution to the problem statement.