Question

For the following regions \(R\), determine which is greater - the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or about the y-axis. \(R\) is bounded by \(y=2 x,\) the \(x\) -axis, and \(x=5\)

Short answer

The solid formed when the region \(R\) is revolved around the y-axis has a larger volume.

Step-by-step solution

Find the volume of the solid when R is revolved around the x-axis

First, we need to find the radius of the solid when revolving the region around the x-axis. The radius of the disk is given by the function y(x) which is \(y = 2x\) in this case. We also need to determine the limits of integration for the x-axis, which are 0 and 5. Now, we can plug these values into the disk method formula and find the volume as follows: $$ V_x = \pi \int_0^5 (2x)^2 dx $$

Calculate the integral for the x-axis revolution

To evaluate the integral, we need to perform the integration itself: $$ V_x = \pi \int_0^5 (4x^2) dx = 4\pi \int_0^5 x^2 dx $$ Integrate and evaluate the definite integral: $$ V_x = 4\pi \left[ \frac{1}{3}x^3 \right]_0^5 = \frac{4\pi}{3}(5^3) = \frac{500\pi}{3} $$ So, the volume of the solid when region \(R\) is revolved around the x-axis is \(\frac{500\pi}{3}\) cubic units.

Find the volume of the solid when R is revolved around the y-axis

Now, we need to find the radius of the solid when the region is revolved around the y-axis. Since the function is given in terms of y, we first need to rewrite the function in terms of x as \(x = \frac{1}{2}y\). The limits of integration for the y-axis are from 0 to 10 since the maximum value of y occurs when x = 5, which gives \(y=2(5)=10\). Now, we can plug these values into the disk method formula and find the volume as follows: $$ V_y = \pi \int_0^{10} (\frac{1}{2}y)^2 dy $$

Calculate the integral for the y-axis revolution

To evaluate the integral, we need to perform the integration itself: $$ V_y = \pi \int_0^{10} (\frac{1}{4}y^2) dy = \frac{\pi}{4} \int_0^{10} y^2 dy $$ Integrate and evaluate the definite integral: $$ V_y = \frac{\pi}{4} \left[ \frac{1}{3}y^3 \right]_0^{10} = \frac{\pi}{12}(10^3) = \frac{1000\pi}{12} $$ So, the volume of the solid when region \(R\) is revolved around the y-axis is \(\frac{1000\pi}{12}\) cubic units.

Compare the volumes

Now that we have the volumes of both solids, we can compare them to determine which one is greater: $$ V_x = \frac{500\pi}{3} \text{ and } V_y = \frac{1000\pi}{12} $$ Comparing these values, we find: $$ \frac{500\pi}{3} < \frac{1000\pi}{12} $$ Therefore, the volume of the solid generated when region \(R\) is revolved around the y-axis is greater than the volume of the solid generated when region \(R\) is revolved around the x-axis.

Key concepts

Disk Method

The disk method is a powerful integration technique used to find the volume of a solid of revolution. It involves "slicing" the solid into very thin disks, each with a small thickness, which is either horizontal or vertical. These disks are stacked up to form the entire solid.

• The radius of each disk is determined by the function that creates the region being revolved. For instance, if a region bounded by the function \( y = f(x) \) is rotated around the x-axis, the radius of each disk is \( f(x) \).
• The thickness of each disk is an infinitesimally small change in the axis perpendicular to the radius. When revolving around the x-axis, the thickness is \( dx \); along the y-axis, it's \( dy \).
• The volume of one disk is given by \( \pi \cdot [\text{radius}]^2 \cdot [\text{thickness}] \).

Calculating the entire volume involves summing the volumes of all these disks through integration.
By applying the disk method formula, \( V = \pi \int [f(x)]^2 \, dx \), you can find the volume of the solid generated by revolving a region around a given axis.

Integral Calculus

Integral calculus is a branch of mathematics focused on accumulation, represented through integration. It helps in finding areas under curves and calculating total quantities like volumes.

• The fundamental components of integral calculus are indefinite and definite integrals. While indefinite integrals provide a general form of antiderivatives, definite integrals result in a numerical value giving the accumulated value between two limits.
• To solve a given integral, you often need to find an antiderivative, which involves reversing differentiation processes following specific rules.
• This discipline is essential for solving real-world problems involving continuous change, such as calculating areas and volumes, or even solving differential equations.

Integral calculus forms the backbone of understanding changes at a gradient level, and its applications extend beyond mathematics into physics, engineering, and statistics.

Definite Integrals

Definite integrals are used to calculate the accumulated quantity over an interval. They are a fundamental concept in calculus and are represented by the integral sign \( \int \) with specified upper and lower limits.

• The key aspect of a definite integral is the limits of integration, denoted as \( \int_a^b \), where \( a \) and \( b \) are specific values marking the interval over which you are integrating. These limits define the stretch between values for calculation.
• The integration process involves finding the antiderivative of a function and then evaluating it from the lower limit to the upper limit using the formula \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of the function.
• Definite integrals give a precise numerical value representing accumulated change from \( a \) to \( b \). They are often employed to compute areas under a curve or solve problems that necessitate summing continuous quantities.

Using definite integrals efficiently helps evaluate physical quantities and other concepts where a precise bounded solution is required, making them indispensable in various fields of study.