Question

(a) Sketch the region \(R\) given by the problem. (b) Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region \(R\) (c) Evaluate one of the iterated integrals to find the signed volume under the surface \(z=f(x, y)\) over the region \(R .\) \(\iint_{R}\left(x^{3} y-x\right) d A,\) where \(R\) is the half of the circle \(x^{2}+y^{2}=9\) in the first and second quadrants.

Short answer

The required integral evaluates to 0.

Step-by-step solution

Understand the Region R

The region \( R \) is the upper half of the circle \( x^2 + y^2 = 9 \). This circle has a radius of 3. Since we consider only the first and second quadrants, \( R \) is defined where \(-3 \leq x \leq 3\) and \(0 \leq y = \sqrt{9 - x^2} \).

Sketch Region R

Draw a circle with a radius of 3, centered at the origin. Highlight the semi-circular part in the first and second quadrants (above the x-axis). This is the region \( R \).

Set Up the Iterated Integral (dy dx)

Consider the integral with \( y \) varying first, from \( 0 \) to \( \sqrt{9-x^2} \), and \( x \) ranging from \(-3 \) to \( 3 \). The iterated integral is: \[ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} (x^3 y - x) \, dy \, dx \]

Set Up the Iterated Integral (dx dy)

Here, \( x \) varies first, from \( -\sqrt{9-y^2} \) to \( \sqrt{9-y^2} \), and \( y \) ranges from \( 0 \) to \( 3 \). The iterated integral is: \[ \int_{0}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} (x^3 y - x) \, dx \, dy \]

Evaluate the Integral (dy dx)

We will solve: \[ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} (x^3 y - x) \, dy \, dx \] Evaluating the inner integral, \[ \int_{0}^{\sqrt{9-x^2}} (x^3 y - x) \, dy = \left[ \frac{x^3 y^2}{2} - xy \right]_{0}^{\sqrt{9-x^2}} \]Simplifying, \[ = \frac{x^3 (9-x^2)}{2} - x\sqrt{9-x^2} \]Therefore, the outer integral becomes: \[ \int_{-3}^{3} \left( \frac{x^3 (9-x^2)}{2} - x\sqrt{9-x^2} \right) \, dx \]

Simplify and Calculate Value

Simplify the expression and evaluate over \( -3 \leq x \leq 3 \): \[ = \int_{-3}^{3} \left( \frac{9x^3}{2} - \frac{x^5}{2} - x\sqrt{9-x^2} \right) \, dx \]Solving involves finding the antiderivatives:\[ \left[ \frac{9x^4}{8} - \frac{x^6}{12} - \left( \int x\sqrt{9-x^2} \, dx \right) \right]_{-3}^{3} \]The integral involving square root can be simplified using trigonometric substitution or symmetry in integrals.

Key concepts

Iterated Integrals

Iterated integrals are used to compute double integrals by breaking them down into a nested sequence of single integrals. This involves integrating a function of two variables over a region in the plane. For a double integral, we can perform integration in two different orders: either with respect to \( y \) first (dy dx), or with respect to \( x \) first (dx dy). In the given exercise, we evaluate the double integral over a region \( R \) by determining these limits of integration for each variable:

• (dy dx): Start by integrating \( y \) from \( 0 \) to \( \sqrt{9-x^2} \), for each \( x \) ranging from \(-3 \) to \( 3 \).
• (dx dy): Alternatively, integrate \( x \) from \(-\sqrt{9-y^2} \) to \( \sqrt{9-y^2} \), while \( y \) varies from \( 0 \) to \( 3 \).

Iterated integrals require careful attention to the limits of integration which describe the region of interest. These limits reflect the boundaries of \( R \) and ensure the function is precisely integrated over the specified region.

Region Sketching

Region sketching is a crucial step in solving double integrals as it helps to visualize the area being studied. For the provided exercise, the region \( R \) is defined as the half of the circle:
\( x^2 + y^2 = 9 \), which is the upper half due to its location in the first and second quadrants of the Cartesian plane. This circle has a center at the origin \( (0,0) \) and a radius of 3.

• Circle Properties: By understanding the equation of a circle, we can easily find its radius and location on the plane. Here, the radius \( r \) is the square root of 9, which equals 3.
• Defining \( R \): Since \( R \) is the upper half, it is crucial to translate this into the limits of our integrals, where \( x \) ranges from \(-3 \) to \( 3 \) and \( y \) goes from \( 0 \) to \( \sqrt{9 - x^2} \).

These sketches provide a logical map that informs how iterated integrals are set up, ensuring that calculations include all necessary parts of the region.

Trigonometric Substitution

Trigonometric substitution is a technique often applied in calculus to simplify integrals involving square roots. This method uses trigonometric identities to transform a difficult integral into a more manageable form. In the exercise at hand, trigonometric substitution can be helpful to solve the integral \(\int x\sqrt{9-x^2} \, dx\).

• Basic Principle: By using a substitution based on the identity \( sin^2\theta + cos^2\theta = 1 \), expressions involving \( \sqrt{a^2-x^2} \) can be rewritten using trigonometric functions.
• Substitution Example: For the expression \( \sqrt{9-x^2} \), a common substitution would be \( x = 3\sin\theta \), leading to \( \sqrt{9-x^2} = 3\cos\theta \).

This process changes the integral into a trigonometric form which can be easier to evaluate. Using appropriate trigonometric identities, such integrals can be worked through to find a solution.

Antiderivatives

In calculus, finding antiderivatives (or indefinite integrals) is essential for solving integrals. An antiderivative for a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). The power of antiderivatives is seen distinctly while evaluating double integrals.

• Applying Antiderivatives: In the given problem, after setting up the integral, the task involves integrating expressions like \( \frac{x^3 y^2}{2} - xy \). Finding antiderivatives for these expressions involves the application of basic integral rules.
• Integral Simplification: Simplifying involves breaking down the components into simpler expressions for which antiderivatives are known, and ensuring that changes of variables or integration limits are correctly applied.

Using proper techniques to find antiderivatives is key in proceeding from the setup of iterated integrals to their actual evaluation and finding the numerical solutions.