Problem 43 - Calculus Single Variable

Question

In each of Exercises 43-48, use the method of cylindrical shells to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region that is bounded by the curve \(y=x^{2}\) and the lines \(y=1, y=4,\) and \(x=3 .\)

Step-by-Step Solution

Answer

The volume of the solid is \\(20.2\pi\\) cubic units.

Step 1: Understand the Problem

We are asked to find the volume of the solid formed by rotating the region \(\mathcal{R}\) around the \(x\)-axis. The region \(\mathcal{R}\) is bounded by the parabola \(y=x^2\), the horizontal lines \(y=1\) and \(y=4\), and the vertical line \(x=3\).

Step 2: Set Up the Integral for Volume using Cylindrical Shells

The formula for the volume using cylindrical shells when rotating about the \(x\)-axis is given by: \[ V = 2\pi \int_{a}^{b} y \cdot (f(y)-g(y)) \; dy \]Where \(f(y)\) and \(g(y)\) are the expressions for \(x\) given \(y\). Here, \(x_1 = \sqrt{y}\) (from \(y=x^2\)) and \(x_2 = 3\). Our bounds for \[y\] are \(1\) and \(4\).

Step 3: Determine the Functions for x in terms of y

For the bounding parabola \(y=x^2\), we solve for \(x\) to get \(x=\sqrt{y}\). So the values of \(x\) in terms of \(y\) for our region are from \(\sqrt{y}\) to \(3\).

Step 4: Write the Volume Integral

Substitute these expressions into the integral:\[ V = 2\pi \int_{1}^{4} y \cdot (3 - \sqrt{y}) \; dy \]This represents the volume of cylindrical shells formed by the region.

Step 5: Integrate

Expand and integrate the expression:\[ V = 2\pi \left( \int_{1}^{4} 3y \; dy - \int_{1}^{4} y^{3/2} \; dy \right) \]The first integral results in \[ 3\cdot \frac{y^2}{2} \bigg|_{1}^{4} = \frac{3}{2} [16 - 1] = \frac{3}{2} \cdot 15 = 22.5\]The second integral results in \[ \frac{2}{5}y^{5/2} \bigg|_{1}^{4} = \frac{2}{5} [32 - 1] = \frac{2}{5} \cdot 31 = 12.4\]

Step 6: Calculate the Final Volume

Subtract the results from the integrals and multiply by \(2\pi\):\[ V = 2\pi (22.5 - 12.4) = 2\pi \cdot 10.1 = 20.2\pi \] Thus, the volume of the solid is \(20.2\pi\) cubic units.

Key Concepts

volume of solids

When we talk about the volume of solids in calculus, we're looking at how much 'space' a 3D object takes up. In problems like this, we're not dealing with just any solid, but specifically those formed by rotating a 2D shape around a line (an axis). This method is effectively used to calculate the volume of the resulting solid.

Imagine drawing a shape on a piece of paper, then rotating that shape around a fixed line, like a spinning top. The 3D shape you get from this rotation is what we find the volume for.

• Volume ensures we understand the size inside a shape.
• It's measured in cubic units.
• Using integrals, we can calculate these volumes even for complex shapes by slicing them into simpler pieces.

Understanding how to apply concepts of volume using calculus techniques like cylindrical shells is crucial for solving these types of problems.

integrals

Integrals are a fundamental concept in calculus used to calculate areas, volumes, and other quantities that add up small pieces of a whole. When dealing with volumes, integrals help us sum up all the infinitesimally small parts to find a total amount.

In the context of our problem, we use integrals to add up the volumes of infinitely thin cylindrical shells, each corresponding to a small slice of the region we are analyzing.

• Think of integral as a sophisticated form of adding up a lot of tiny pieces.
• We use the boundaries of the region we're interested in as the limits of integration.
• Integrals can be tricky, but they are just a consistent way to sum areas or volumes.

Mastering integrals opens up a world of possibilities in analyzing complex spatial dimensions like the volumes of solids.

rotation about the x-axis

Rotation about the x-axis is a method used to transform a 2D region into a 3D shape. It's a specific type of rotation where the x-axis is the line around which everything spins.

By visualizing this, any point on the 2D shape will circle around, tracing out a 3D form—a solid of revolution.

• This rotational movement is similar to how wheels turn around an axle.
• Every point on the curve traces a circular path, contributing to the full 3D shape.
• It involves understanding the geometric movement to apply correct formulas.

This concept is important for forming the shape, making techniques like cylindrical shells applicable for calculating volume.

parabolas

Parabolas are key shapes in algebra and calculus, often appearing when dealing with quadratic functions like the one here: \( y = x^2 \).

They have a characteristic U-shape, and understanding how they behave is essential when calculating areas and volumes, as in this exercise.

• A parabola can open upwards or sideways, but in this exercise, it opens vertically.
• The shape is symmetric around its axis of symmetry – a vertical line down its center.
• Parabolas are essential in defining regions in integral problems.

In our example, the parabola provides one boundary of the planar region, affecting how the solid shape is formed during rotation around the x-axis.