Problem 51 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R} 3 x^{2} d A ; R\) is bounded by \(y=0, y=2 x+4,\) and \(y=x^{3}\).

Step-by-Step Solution

Answer

Answer: The approximate value of the double integral is 11.063.

Step 1: Sketch the region R

First, let's plot the three given equations on a 2D plane: y = 0 is the x-axis, y = 2x + 4 is a straight line with slope 2 and intercept 4, and y = x^3 is a cubic curve. By graphing these functions, we can see that the region R is a closed area bounded by these equations.

Step 2: Find the intersections of the curves

We need to find the points where the curves intersect to determine the limits of integration. To find the intersection between y = 2x + 4 and y = x^3, set the two equations equal to each other: x^3 = 2x + 4 Rearranging the equation and setting it to zero, we get: x^3 - 2x - 4 = 0 We can numerically solve this equation to find the intersection point x ≈ 1.587. Thus, the intersection point is approximately (1.587, 7.174). To find the intersection between y = 0 and y = x^3, set the two equations equal to each other: x^3 = 0 The only intersection point is x = 0, and thus the point is (0,0). Similarly, for y = 0 and y = 2x + 4: 2x + 4 = 0 Solving for x, we get x = -2, thus the point is (-2,0).

Step 3: Set up the limits of integration

We need to express the limits of integration in terms of x and y. Since the region R is comfortably expressed in terms of x, we will integrate with respect to y first and then x. The limits of integration with respect to y are from y = 0 to y = x^3, and with respect to x are from x = -2 to x = 1.587. Now we can set up the double integral: \(\int_{-2}^{1.587}\int_{0}^{x^3} 3x^2\, dy\, dx\)

Step 4: Evaluate the double integral

First, we integrate with respect to y: \(\int_{-2}^{1.587} [3x^2y]_{0}^{x^3} dx\) This simplifies to: \(\int_{-2}^{1.587} 3x^2(x^3) dx\) Next, we integrate with respect to x: \(\left[\frac{3}{6}x^6\right]_{-2}^{1.587}\)

Step 5: Calculate the final result

Now we evaluate the integral to get our final result: \(\frac{1}{2}[(1.587)^6 - (-2)^6] \approx 11.063\) Thus, the value of the double integral is approximately 11.063.

Key Concepts

Region of Integration

When dealing with double integrals, understanding the region of integration, often represented as 'R', is essential. This region is the specific area over which we integrate a given function. In our problem, the region R is bounded by the following curves:

• The x-axis, represented by the line \( y = 0 \)
• The line \( y = 2x + 4 \)
• The curve \( y = x^3 \)

Visualizing these curves on a 2D graph helps us see that they enclose a specific region. Finding this region involves determining where these curves intersect, which provides the essential boundaries or limits of our integration. By plotting these lines and curves, we ensure that we are integrating over the correct area. Sketching these relationships gives a clear depiction of where the curves cut each other leading to a closed and defined region.

Limits of Integration

The limits of integration delineate the span over which we compute the integral. It's crucial to find these limits correctly to evaluate the integral accurately. In our problem, the limits come from the intersection points of the given curves.

• For the line \( y = 2x + 4 \) and the curve \( y = x^3 \), we solved the equation \( x^3 = 2x + 4 \), resulting in an intersection point at approximately \( x = 1.587 \).
• For \( y = 0 \) intersecting with both \( y = x^3 \) and \( y = 2x + 4 \):
  • The intersection with \( y = x^3 \) is at \( x = 0 \).
  • For \( y = 2x + 4 \), solving gives the intersection point \( x = -2 \).

Thus, for our double integral, the limits are set as: the inner integral runs from \( y = 0 \) to \( y = x^3 \), and the outer integral runs from \( x = -2 \) to \( x = 1.587 \). These boundaries encapsulate the region of integration necessary for accurate computation.

Numerical Solution

In cases where intersection of functions yields complex equations, sometimes numerical solutions are an efficient path to finding exact or approximate values. For the intersection \( x^3 = 2x + 4 \), the analytical solution may not be straightforward or could be cumbersome.To solve \( x^3 - 2x - 4 = 0 \) numerically, techniques such as the bisection method, Newton's method, or other root-finding algorithms are employed.

• These methods involve iterative procedures that hone in on precise x-values, in this case resulting in \( x \approx 1.587 \).

Such numerical methods are important tools in calculus and engineering, especially when handling equations that are non-trivial to solve algebraically. They provide us with acceptable levels of precision, ensuring that we can proceed with the integral evaluation accurately.

Cubic Equation

Cubic equations are polynomial equations of degree three, and in our exercise, we encountered \( y = x^3 \) as a boundary of the region.

• This curve forms part of the integration boundary and intersects with lines to define limits.
• Generally, the cubic equation \( ax^3 + bx^2 + cx + d = 0 \) has coefficients that determine its shape and position on a graph.

Understanding this type of equation involves recognizing its characteristic features:

• One real root or up to three distinct real roots, depending on coefficients.
• Its graph is typically a smooth, continuous curve with possible inflection points.

The manipulation and solving of cubic equations, either graphically or analytically, are common in calculus. They are critical for determining where curves intersect with other functions, thus helping to find integration regions exactly as needed for accurately computing double integrals.