Problem 58 - Thomas Calculus

Question

Find the volume of the solid that is bounded above by the cylinder \(z=x^{2}\) and below by the region enclosed by the parabola \(y=2-x^{2}\) and the line \(y=x\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{63}{20} \) cubic units.

Step 1: Set Up the Boundaries

Understand the region over which we are integrating. The solid is between the curve \[ y = 2 - x^2 \] and the line \[ y = x \] in the xy-plane. Set them equal to find points of intersection:\[ 2 - x^2 = x \]\[ x^2 + x - 2 = 0 \]Factor this quadratic:\[ (x - 1)(x + 2) = 0 \]The solutions are \( x = 1 \) and \( x = -2 \). The region is between these x-values.

Step 2: Set Up the Integral for Volume Calculation

The volume is given by the integral of the height of the solid over the region in the xy-plane. The height is determined by the cylinder and thus we integrate the cylinder function:\[ z = x^2 \]The integration is with respect to \( x \) from -2 to 1, and with respect to \( y \) from the line \( y = x \) (lower limit) to the parabola \( y = 2 - x^2 \) (upper limit).

Step 3: Write the Double Integral

The double integral is used to calculate the volume:\[ V = \int_{-2}^{1} \int_{x}^{2 - x^2} x^2 \, dy \, dx \] This represents the volume under the surface \( z = x^2 \) over the region bounded by the curves \( y = x \) and \( y = 2-x^2 \).

Step 4: Perform the Inner Integration

Integrate with respect to \( y \) first:\[ \int_{x}^{2-x^2} x^2 \, dy = x^2 \cdot (y \bigg|_{x}^{2-x^2}) \]\[ = x^2 \cdot [(2-x^2) - x] = x^2 \cdot (2-x^2 - x) \]\[ = x^2 \cdot (2-x-x^2) = x^2 \cdot (2-x-x^2) \]

Step 5: Simplify and Prepare for Outer Integration

Simplify the expression for the inner integral:\[ = x^2 \cdot (2 - x - x^2) \]Expand this:\[ = 2x^2 - x^3 - x^4 \]

Step 6: Perform the Outer Integration

Integrate the result from the last step with respect to \( x \) from -2 to 1:\[ \int_{-2}^{1} (2x^2 - x^3 - x^4) \, dx \]\[ = \left[ \frac{2x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} \right]_{-2}^{1} \]

Step 7: Evaluate the Definite Integral

Substitute the limits into the antiderivative:For \( x = 1 \):\[ \frac{2(1)^3}{3} - \frac{(1)^4}{4} - \frac{(1)^5}{5} = \frac{2}{3} - \frac{1}{4} - \frac{1}{5} \]Calculate the fractions:\[ = \frac{40}{60} - \frac{15}{60} - \frac{12}{60} = \frac{13}{60} \]For \( x = -2 \):\[ \frac{2(-2)^3}{3} - \frac{(-2)^4}{4} - \frac{(-2)^5}{5} \]\[ = \frac{-16}{3} - 4 + \frac{32}{5} \]Convert to a common denominator:\[ = \frac{-80}{15} - \frac{60}{15} + \frac{96}{15} = \frac{-44}{15} \]Subtract the values:\[ V = \left( \frac{13}{60} \right) - \left( \frac{-44}{15} \right) \]Find a common denominator and compute:\[ V = \frac{13}{60} + \frac{176}{60} = \frac{189}{60} = \frac{63}{20} \]

Step 8: Final Step: State the Volume

The volume of the solid bounded by the given surfaces and curves is:\( \frac{63}{20} \) cubic units.

Key Concepts

Volume Calculation

In this problem, our goal is to find the volume of a solid bounded by certain surfaces. The volume of this solid can be found by using calculus, specifically the double integral. To begin, you need to understand that the double integral computes the volume under a surface over a specific region in the xy-plane.

Think of the xy-plane as a piece of paper, and the surface we are integrating under is like a roof over that paper. The volume calculation involves finding the total volume of space between the roof and the paper (defined region). The height of the solid, or the distance between the surfaces, is determined by the equation of the higher surface, in this case, the cylinder given by the equation \( z = x^2 \).

• Identify the surface equation. Here, it is \( z = x^2 \).
• Determine the region over which the integral is evaluated.
• The result of the double integral gives the volume of the solid.

Bounded Region

Understanding the bounded region is crucial for setting up integrals. Here, we need to find the area enclosed between two curves in the xy-plane: the parabola \( y = 2 - x^2 \) and the line \( y = x \). These curves define the limits for our integral.

• Find the intersection points of the curves to know the limits of integration along the x-axis. Solve \( 2 - x^2 = x \) to get the intersection points \( x = 1 \) and \( x = -2 \).
• This interval \([-2, 1]\) represents the x-values over which we will calculate the volume.
• The y-values vary from the line \( y = x \) (lower limit) to the parabola \( y = 2 - x^2 \) (upper limit).

So, the integral is set up properly to account for these boundaries from one function to the other, forming a closed region in the plane.

Cylindrical Coordinates

While the problem here uses rectangular coordinates, it's valuable to understand the role of cylindrical coordinates in similar contexts. Cylindrical coordinates, generally denoted as \((r, \theta, z)\), are immensely useful for problems involving symmetry about an axis, like cylinders or surfaces defined radially.

The key formula transformations from rectangular to cylindrical coordinates involve:

• \(x = r \cos \theta\),
• \(y = r \sin \theta\),
• \(z = z\).

Such transformations simplify solving some integrals, especially those involving circular symmetries, which are complex in rectangular coordinates. Even though this problem doesn't explicitly use cylindrical coordinates, having this knowledge helps solve other problems involving similar volumes and surfaces.

Intersection Points

Intersection points are the coordinates where two or more curves meet. They are essential for defining integration limits, helping us set up the region we integrate over. You begin by setting the equations of the curves to each other. For our problem, the curves are \( y = 2 - x^2 \) and \( y = x \).

Setting these equations equal gives \( 2 - x^2 = x \). Solving this quadratic equation allows you to find the x-values where the intersection happens. Here, factored as \( (x - 1)(x + 2) = 0 \), reveals intersections at \( x = 1 \) and \( x = -2 \).

• These points, \( x = 1 \) and \( x = -2 \), thus define the region of integration with respect to x.
• Use intersection points to form the limits for setting up the integral.
• They ensure that the integration covers the entire bounded region correctly.

Finding these points are pivotal in ensuring that your integration accurately reflects the area over which the volume is calculated.