Question

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(0

Short answer

The volume of the solid generated when R is revolved around the x-axis is finite if p < 0. Additionally, the volume of the solid generated when R is revolved around the y-axis is finite if p > 0.

Step-by-step solution

Setting Up the Integral for Revolution about the \(x\)-axis

We are given that the region \(R\) is bounded by the graph of the function \(f(x) = x^{-p}\) and the \(x\)-axis for \(0 < x \leq 1\). When revolved around the \(x\)-axis, we create a solid of revolution. To find the volume of the solid, we can use the Disk Method: $$V_x = \pi \int_{0}^{1} f(x)^2 dx = \pi \int_{0}^{1} (x^{-p})^2 dx$$

Evaluating the Integral for Revolution about the \(x\)-axis

Now we evaluate the integral: $$V_x = \pi \int_{0}^{1} x^{-2p} dx$$ Now find the antiderivative of \(x^{-2p}\): $$V_x = \frac{\pi}{-2p + 1} [x^{-2p + 1}]_{0}^{1}$$ Here, we need to find values of \(p\) for which the integral converges, i.e., the volume is finite.

Determine the Values of \(p\) for which the Integral Converges

The integral converges if \(-2p+1 > 1\), which implies: $$-2p > 0 \Rightarrow p < 0$$ Now, let's solve part (b).

Find the Inverse Function

To set up the integral for revolution around the \(y\)-axis, we first need to find the inverse of \(f(x) = x^{-p}\). Solving for \(x\) in terms of \(y\), the inverse function is given by: $$x = y^{-\frac{1}{p}}$$

Setting Up the Integral for Revolution about the \(y\)-axis

Now, we can set up the integral using the Disk Method with respect to \(y\): $$V_y = \pi \int_{0}^{1} (y^{-\frac{1}{p}})^2 dy = \pi \int_{0}^{1} y^{-\frac{2}{p}} dy$$

Evaluating the Integral for Revolution about the \(y\)-axis

Now we evaluate the integral: $$V_y = \pi \int_{0}^{1} y^{-\frac{2}{p}} dy$$ Now find the antiderivative of \(y^{-\frac{2}{p}}\): $$V_y = \frac{\pi}{\frac{2}{p} + 1} [y^{\frac{2}{p} + 1}]_{0}^{1}$$

Determine the Values of \(p\) for which the Integral Converges

The integral converges if \(\frac{2}{p} + 1 > 1\), which implies: $$\frac{2}{p} > 0 \Rightarrow p > 0$$ In conclusion: a. For part (a), the volume of the solid generated when \(R\) is revolved about the \(x\)-axis is finite if \(p < 0\). b. For part (b), the volume of the solid generated when \(R\) is revolved about the \(y\)-axis is finite if \(p > 0\).

Key concepts

Solid of Revolution

In calculus, a solid of revolution is a three-dimensional shape formed by rotating a two-dimensional area or region around an axis. This is like spinning a flat shape around a line to create a 3D object. Imagine taking a semicircle and rotating it around its diameter: you'd get a sphere! The concept is widely used in practical applications, such as determining the shape of objects like bottles, lampshades, or even engine parts.

The idea is fundamental to understanding how objects we see in everyday life can be envisioned and analyzed mathematically. When discussing solids of revolution, always consider which axis is used for the rotation, as it greatly determines the outcome of the shape.

In the context of the given exercise, rotating the function defined by the graph of the equation around the x-axis or y-axis forms different solids of revolution, and determining the volume is a critical step to understanding the properties of these solids.

Disk Method

The Disk Method is a mathematical technique used to find the volume of a solid of revolution. This method specifically simplifies volume calculation when an area is revolved around an axis, forming disk-shaped cross-sections perpendicular to the axis of rotation. Each disk's volume can be thought of like a thin, flat cylinder.

To apply the Disk Method, imagine slicing the solid of revolution into a series of thin disks. The formula involves setting up an integral to sum up the volumes of these disks across a range. For a solid revolving around the x-axis, you use the formula:

\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]

Here, each disk has a volume of \( \pi [f(x)]^2 \Delta x \), where \( f(x) \) is the radius of the disk. This radius changes as \( x \) changes from \( a \) to \( b \). When this integral is evaluated, the total volume of the solid is obtained.

In the exercise, two scenarios of rotation are evaluated: one around the x-axis and the other about the y-axis, using the disk method to derive formulae for the volumes that change with respect to variable \( p \).

Volume of Solids

The volume of a solid is a measure of the amount of three-dimensional space it occupies. In mathematical terms for solids of revolution, this volume can be computed through integration. Calculating the volume of solids formed by rotating a function around an axis relies heavily on understanding both the limits of integration and the integrand itself, which defines the shape.

The integration process needs careful setup because, for convergence—meaning the result is a finite number—the function and its powers must be manageable within the limits. As shown in the exercise, volume becomes finite under specific conditions for the exponent \( p \).

For rotation around the x-axis, \( V_x = \pi \int_{0}^{1} (x^{-p})^2 dx \) is finite when \( p \) is less than zero, implying the solid formed has a finite size. Conversely, when revolving around the y-axis, the criteria change; \( V_y = \pi \int_{0}^{1} (y^{-rac{2}{p}})^2 dy \) converges to a finite volume when \( p \) is greater than zero. Identifying these conditions is crucial in mathematical exploration to know when the calculations deliver meaningful physical results.