Problem 30 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R} x^{2} y d A ; R\) is the region in quadrants 1 and 4 bounded by the semicircle of radius 4 centered at (0, 0).

Step-by-Step Solution

Answer

Question: Determine the value of the double integral \(\iint_{R} x^{2} y\, dA\), where R is the region in quadrants 1 and 4 bounded by the semicircle of radius 4 centered at the origin. Answer: The value of the given double integral is \(\frac{1024}{15}\).

Step 1: Sketch the region R in the Cartesian plane

To visualize the region R, draw the semicircle of radius 4 centered at the origin (0, 0) in the Cartesian plane. The equation of this semicircle is \(x^2 + y^2 = 4^2\), where y is non-negative in quadrant 1 (y ≥ 0), and y is non-positive in quadrant 4 (y ≤ 0). Mark the region R in quadrants 1 and 4 enclosed by this semicircle.

Step 2: Choose the order of integration and find the limits

Let's integrate first with respect to y and then with respect to x. For a fixed x value in the region R, the y values will range from the bottom of the semicircle to the top of the semicircle. Therefore, the limits of integration for y will be from \(-\sqrt{16-x^2}\) to \(\sqrt{16-x^2}\). The x values will range from the leftmost point of the semicircle to the rightmost point, so the limits of integration for x will be from -4 to 4.

Step 3: Set up the double integral

The double integral of the given function over the region R can be expressed as: $$\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} x^2y\, dy\, dx$$

Step 4: Evaluate the inner integral

To evaluate the inner integral with respect to y, we have: $$\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} x^2y\, dy = \frac{1}{2}x^2y^2\Big|_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} = x^2(\sqrt{16-x^2})^2 - x^2(-\sqrt{16-x^2})^2 = x^2(16-x^2)-x^2(x^2-16) = 32x^2-2x^4$$

Step 5: Evaluate the outer integral

Now, we need to evaluate the outer integral with respect to x: $$\int_{-4}^{4} (32x^2-2x^4) dx = \left[\frac{32}{3}x^3 - \frac{2}{5}x^5\right]_{-4}^4 = \left[\frac{32}{3}(4)^3 - \frac{2}{5}(4)^5\right] - \left[\frac{32}{3}(-4)^3 - \frac{2}{5}(-4)^5\right] = 2\left(\frac{32}{3}(4)^3 - \frac{2}{5}(4)^5\right) = \frac{1024}{15}$$ So, the value of the given double integral is: $$\iint_{R} x^{2} y\, dA = \frac{1024}{15}$$

Key Concepts

Cartesian Plane

The Cartesian Plane, also known as the coordinate plane, is a two-dimensional surface defined by two perpendicular axes: the x-axis (horizontal) and y-axis (vertical). This plane allows us to locate points using pairs of numeric coordinates \((x, y)\), which represent distances from the origin (0, 0).

• The origin divides the plane into four regions called quadrants.
• In applied mathematics, the Cartesian plane is an essential tool for defining geometric shapes and regions of integration.

Understanding the Cartesian plane is crucial when working with double integrals, as it helps visualize the areas over which the integration occurs, like in this exercise with the semicircle located in quadrants 1 and 4.

Semicircle

Geometrically, a semicircle represents half of a circle. In this case, we are looking at a semicircle with a radius of 4 centered at the origin (0, 0). The formula for the full circle is given by \(x^2 + y^2 = 16\). For a semicircle,

• In quadrant 1, where \(y \geq 0\), the curve forms the upper half of the circle.
• In quadrant 4, where \(y \leq 0\), it forms the lower half of the circle.

To understand the region of integration (R) in this exercise, imagine cutting the circle in half horizontally. These two halves encompass the areas where integration is performed, specifically from the bottom to the top of the semicircle for any value along the x-axis within the circle's boundaries.

Order of Integration

The order of integration refers to the sequence in which we perform multiple integrals, often denoted as DXDY or DYDX. In double integrals, this order can be chosen to simplify calculations. Specifically for our problem:

• The choice to integrate first with respect to \(y\) simplifies evaluation, as \(y\) limits depend on \(x\).
• Calculating \(y\) first involves integrating the vertical 'slices' and then accumulating these across the \(x\) range.

Selecting the correct order of integration helps in handling complex regions, potentially reducing the difficulty of solving the integral.

Limits of Integration

The limits of integration determine the bounds over which the integration occurs. These limits must be carefully assessed based on the region R defined. In our scenario:

• For \(x\), the integral bounds range from -4 to 4, covering the entire semicircle along the x-axis.
• For \(y\), the bounds are from \(-\sqrt{16-x^2}\) to \(\sqrt{16-x^2}\), which account for the top and bottom of the semicircle respectively at any given x-coordinate.

Correctly identifying these bounds is crucial for solving double integrals as it ensures the entire specified region is accounted for without overextending beyond its defined space.