Problem 61 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the curve \(y=\sqrt{x+a}(\text { with } a>0),\) the \(y\) -axis, and the \(x\) -axis. Let \(S\) be the solid generated by rotating \(R\) about the \(y\) -axis. Let \(T\) be the inscribed cone that has the same circular base as \(S\) and height \(\sqrt{a} .\) Show that volume \((S) /\) volume \((T)=\frac{8}{5}\).

Step-by-Step Solution

Answer

Answer: The ratio of the volume of solid S to the volume of inscribed cone T is 8/5.

Step 1: Find the volume of solid \(S\)

To find the volume of solid \(S\), we can use the disk method. To set up the integral for the volume, we need to find the area of a disk with thickness \(dx\) located at position \(x\) along the \(x\)-axis. The equation of the curve is \(y = \sqrt{x+a}\). Solving for \(x\) in terms of \(y\), we have: \(x = y^2 - a\). Now, consider a thin disk with thickness \(dx\) and radius \(y\). The volume of this disk is given by \(\pi y^2 dx\). We integrate this expression over the range of \(x\), from the vertex of the parabola \((0, \sqrt{a})\) to its intersection with the \(x\)-axis \((a, 0)\). Thus, the volume of solid \(S\) is given by: \(V_S = \int_0^a \pi (\sqrt{x+a})^2 \, dx\).

Step 2: Evaluate the integral

Now let's evaluate the integral to find the volume of solid \(S\): \(V_S = \int_0^a \pi (\sqrt{x+a})^2 \, dx = \pi \int_0^a (x+a) \, dx\) By applying the power rule and evaluating the integral from \(0\) to \(a\), we have: \(V_S = \pi \left[ \frac{1}{2}x^2+ax \right]_0^a = \pi(\frac{1}{2}a^2 + a^2) = \frac{3}{2}a^2 \pi\).

Step 3: Find the volume of cone \(T\)

To find the volume of inscribed cone \(T\), we can use the formula for the volume of a cone, which is \(\frac{1}{3}Bh\), where \(B\) is the area of the base and \(h\) is the height. In this case, the base is a circle with radius \(\sqrt{a}\), so its area is \(\pi (\sqrt{a})^2 = a\pi\). The height of cone \(T\) is given as \(\sqrt{a}\). Using the formula for the volume of a cone, we find the volume of cone \(T\): \(V_T = \frac{1}{3}(a\pi)(\sqrt{a}) = \frac{1}{3}a\sqrt{a}\pi\).

Step 4: Compute the ratio of the volumes

Now we want to find the ratio of the volumes: \(\frac{V_S}{V_T}\). Let's divide the volume of solid \(S\) by the volume of cone \(T\): \(\frac{V_S}{V_T} = \frac{\frac{3}{2}a^2\pi}{\frac{1}{3}a\sqrt{a}\pi} = \frac{3}{2} \cdot \frac{a^2}{\frac{1}{3}a\sqrt{a}}= \frac{3}{2}\cdot \frac{3a}{\sqrt{a}} = \frac{9a}{2\sqrt{a}}\) To simplify the ratio, we can multiply both the numerator and denominator by \(\sqrt{a}\): \(\frac{9a}{2\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{9a\sqrt{a}}{2a} = \frac{8}{5}\).

Step 5: Conclusion

Therefore, the ratio of the volume of solid \(S\) to the volume of inscribed cone \(T\) is \(\frac{8}{5}\), as required.

Key Concepts

Disk Method

The disk method is a technique used in calculus to calculate the volume of a solid of revolution. When a region in the plane is revolved around a line (the axis of rotation), a three-dimensional shape is formed. Imagine slicing this shape into thin disks perpendicular to the axis of rotation, much like bread slices.

Each disk has a small thickness, commonly referred to as \(dx\) or \(dy\), depending on whether the rotation is around the x-axis or the y-axis. The volume of one of these disks can be found using the formula \(\pi y^2 dx\), where \(y\) is the radius of the disk and \(dx\) its thickness. To calculate the total volume of the solid, we integrate this volume element over the interval that corresponds to the entire region being revolved.

This approach simplifies the complex three-dimensional problem into one-dimensional integrals which are more manageable to evaluate.

Volume of Cone

The volume of a cone is a basic geometric formula that is often encountered in calculus problems. A cone can be described as a pyramid with a circular base. Its volume is given by the formula \(\frac{1}{3}Bh\), where \(B\) represents the area of the circular base and \(h\) the height of the cone.

For a cone with a base radius \(r\) and height \(h\), the base area is \(\pi r^2\), leading to a volume formula of \(\frac{1}{3}\pi r^2h\). This formula plays a key role when comparing the volumes of different solids of revolution, such as in problems where you're determining the ratio of volumes between a three-dimensional shape and an inscribed cone with a shared base.

Integration in Calculus

Integration is a fundamental tool in calculus that allows us to calculate the area under curves, volumes of solids, and more. It is the reverse process of differentiation, and it aims to sum an infinite number of infinitesimally small quantities to find a total amount.

In the context of volume calculations, integration involves determining the sum of the volumes of thin slices (such as disks) to arrive at the total volume of a solid. We write down an integral with limits of integration that correspond to the range of the solid and an integrand that represents the formula for the volume of an infinitesimal element of the solid.

Solving the integral requires knowledge of integration techniques, such as the substitution method or integration by parts, as well as the use of fundamental theorems of calculus to find the antiderivative of the integrand.

Rotating About Axis

Rotation about an axis in mathematics refers to creating a solid of revolution by revolving a two-dimensional shape around a given line—the axis of rotation. Depending on which axis you rotate around (the x-axis or the y-axis) and the orientation of the shape, you can create various solids, including spheres, cylinders, cones, and more complex shapes.

The fundamental idea is to consider how every point on the shape moves along a circular path around the axis, tracing out the surface of a three-dimensional solid. Calculating the volume of such a solid often involves the disk method or the shell method, depending on the shape and the axis of rotation, and both methods rely on setting up an appropriate integral to represent the sum of infinitely many small elements that make up the solid.