Problem 59 - Thomas Calculus

Question

Find the volume of the solid whose base is the region in the \(x y\) -plane that is bounded by the parabola \(y=4-x^{2}\) and the line \(y=3 x,\) while the top of the solid is bounded by the plane \(z=x+4\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{171}{4}\).

Step 1: Sketch the Region

Visualize the area in the xy-plane bounded by the curves \(y = 4 - x^2\) and \(y = 3x\). These two curves intersect, which will help define the limits of integration. Let us set \(4 - x^2 = 3x\) to find points of intersection.

Step 2: Find Points of Intersection

Set the two equations equal: \(4 - x^2 = 3x\). Rearranging gives us \(-x^2 - 3x + 4 = 0\). Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -1, b = -3, c = 4\). This gives intersection points as \(x = -4\) and \(x = 1\).

Step 3: Establish the Volume Integral Setup

The volume of the solid is the double integral of \(z\) over the bounded region \(R\) in the xy-plane. The top of the solid is the plane \(z = x + 4\). The height function is \(h(x,y) = x + 4\) over the region \(R: 3x \leq y \leq 4 - x^2\), with \(x\) ranging from \(-4\) to \(1\).

Step 4: Set Up the Double Integral

The volume \(V\) can be computed by the double integral \(\int_{-4}^{1} \int_{3x}^{4-x^2} (x + 4) \, dy \, dx\). This integrates \(z = x + 4\) over the region, first integrating with respect to \(y\) from \(3x\) to \(4 - x^2\).

Step 5: Solve the Inner Integral

Evaluate the inner integral \(\int_{3x}^{4-x^2} (x + 4) \, dy\). This simplifies to \[(x+4)(y) \bigg|_{3x}^{4-x^2} = (x+4)(4-x^2) - (x+4)(3x)\].

Step 6: Simplify and Evaluate Outer Integral

Simplify the expression: \[(x+4)(4-x^2) - (x+4)(3x) = (4x + 16 - x^3 - 4x^2) - (3x^2 + 12x)\] \[= -x^3 - 7x^2 - 8x + 16\]. Integrate: \[\int_{-4}^{1} (-x^3 - 7x^2 - 8x + 16) \, dx\].

Step 7: Solve the Outer Integral

Compute: \[\int (-x^3 - 7x^2 - 8x + 16) \, dx = -\frac{x^4}{4} - \frac{7x^3}{3} - 4x^2 + 16x \bigg|_{-4}^{1}\]. Evaluating this from \(-4\) to \(1\) gives the volume.

Step 8: Calculate Final Volume

Substitute the boundaries into the antiderivative: \[-\left(\frac{1}{4}\right) - \left(\frac{7(1)^3}{3}\right) - 4(1)^2 + 16(1) + (\left(\frac{256}{4}\right) + \left(\frac{7\cdot 64}{3}\right) + 64 + 64)\]; Simplify to find the final volume.

Key Concepts

Double Integral

When finding the volume of a solid, especially one bounded by various curves or surfaces, double integrals come into play. A double integral essentially computes the accumulation of a quantity, such as volume, over a two-dimensional area. In this case, we have a region in the xy-plane that will be used to determine how high
each point stacks up to create the solid.

• To use a double integral, we first need to know the bounds of our region of integration.
• The integrand is the function that gives the height of the solid above each point of the region.

In our problem, the height is defined by the plane equation given. The double integral accumulates these individual volumes across the entire region. Once set up, the double integral is evaluated iteratively: first, integrate over one variable while treating the other as constant, then repeat for the remaining variable.

Parabola and Line Intersection

Finding the intersection points of a parabola and a line is key in solving this exercise. These intersections define the limits of integration in one direction when using a double integral. Here, the parabola is described by the equation:
\( y = 4 - x^2 \)
and the line by\( y = 3x \).

• To find where they intersect, set these equations equal: \( 4 - x^2 = 3x \).
• Rearrange this to form a quadratic equation: \( -x^2 - 3x + 4 = 0 \).

You solve this quadratic equation using the quadratic formula. The solutions provide the x-values of the intersection points, which here are \( x = -4 \) and \( x = 1 \). These points delineate where the curves cross, effectively framing the width of the region of interest in the xy-plane.

Limits of Integration

Defining the limits of integration is crucial for solving a double integral. These limits specify the region over which the function is to be integrated.

• The x-values from the intersection points \(-4\) and \(1\) are used as the limits for outer integration.
• The inner limits are determined by the curves themselves: for each specific x, y ranges from \(3x\) to \(4 - x^2\).

These combined limits effectively slice the solid into infinitesimally thin strips, parallel to the y-axis. By integrating first over y and then over x, you cover all the strips that make up the solid. This method ensures that every point within the defined boundary is accounted for in the volume calculation.

Height Function

The height function in this volume problem is provided by the plane equation \( z = x + 4 \). This function determines the vertical distance from the base region in the xy-plane to the top of the solid.

• The plane's equation gives a height for every point (x, y) in the region.
• When you graph this in 3D, each point in the xy-region corresponds to a point directly above it in the z-dimension.

The height function is crucial because it signifies how high each individual segment contributes to the overall volume. The double integral sums up these contributions to give the entire volume of the solid above the xy-plane region defined by the original curves. This composite structure integrates the height function over all valid (x, y) pairs defined by our region of interest.