Problem 36
Question
Find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region and a typical disk element. \(y=\sqrt{2 x}, y=0, x=2\)
Step-by-Step Solution
Verified Answer
The volume is \( 4\pi \) cubic units.
1Step 1: Understand the Problem
We need to find the volume of the solid generated by rotating the region under the curve \( y = \sqrt{2x} \), from \( x = 0 \) to \( x = 2 \), around the \( x \)-axis. The curve \( y = \sqrt{2x} \) and the line \( y = 0 \) form the upper and lower boundaries, respectively, while \( x = 2 \) is the vertical line cutting the area.
2Step 2: Identify the Disk Element
For the disk method, the typical disk has radius \( R = \sqrt{2x} \) and thickness \( dx \). The area of the disk is \( A = \pi R^2 = \pi (\sqrt{2x})^2 = 2\pi x \).
3Step 3: Set up the Integral
To calculate the volume, integrate the area of the disks along the \( x \)-axis from 0 to 2. The volume \( V \) is given by the integral \( V = \int_{0}^{2} 2\pi x \, dx \).
4Step 4: Solve the Integral
Compute the integral \( \int_{0}^{2} 2\pi x \, dx \). This simplifies to \( 2\pi \int_{0}^{2} x \, dx \). Evaluating this gives:\[ 2\pi \left[ \frac{x^2}{2} \right]_{0}^{2} = 2\pi \times \frac{2^2}{2} - 2\pi \times \frac{0^2}{2} = 2\pi \times 2 = 4\pi. \]
5Step 5: Conclusion
The volume of the solid generated by rotating the region bounded by the curves is \( 4\pi \) cubic units.
Key Concepts
Disk MethodIntegral CalculusRotating Around the x-axis
Disk Method
The disk method is a valuable technique in calculus used to find the volume of a solid of revolution. It involves slicing the solid into thin, circular disks, each parallel to the axis of rotation, and then integrating the volume of each disk to find the total volume. Think of a stack of coins; each "coin" represents a disk in your solid.
To apply the disk method, follow these steps:
To apply the disk method, follow these steps:
- Identify the region of interest that will be rotated.
- Choose the axis of rotation (usually, this is the x-axis or y-axis).
- Determine the radius of each disk; often this is the function value at a given point.
- Decide the thickness of each disk, typically an infinitesimally small increment, represented as dx or dy.
Integral Calculus
Integral calculus is a branch of mathematics focused on integrals. It plays a crucial role in determining volumes, areas, and other cumulative quantities. The integral represents the accumulation of tiny quantities over an interval.
When considering volumes of solids of revolution, we rely on the integral to sum up the infinitesimally small components, such as the disk areas. The integral for the disk method is typically set up as \( V = \int_a^b \pi R(x)^2 \, dx \), where \( a \) and \( b \) define the interval and \( R(x) \) is the radius.
Solving an integral involves finding the antiderivative of the function to be integrated and then evaluating this antiderivative at the boundaries of the interval. In the volume problem provided, this translates into converting a complicated geometric problem into a manageable algebraic one.
When considering volumes of solids of revolution, we rely on the integral to sum up the infinitesimally small components, such as the disk areas. The integral for the disk method is typically set up as \( V = \int_a^b \pi R(x)^2 \, dx \), where \( a \) and \( b \) define the interval and \( R(x) \) is the radius.
Solving an integral involves finding the antiderivative of the function to be integrated and then evaluating this antiderivative at the boundaries of the interval. In the volume problem provided, this translates into converting a complicated geometric problem into a manageable algebraic one.
Rotating Around the x-axis
When we talk about rotating a region around the x-axis, it means that we take a two-dimensional area and spin it around the x-axis to create a three-dimensional solid.
The method of rotation influences the shape and size of the resulting solid. In the exercise, the area under the curve \( y = \sqrt{2x} \) between \( x = 0 \) and \( x = 2 \) is "swept" around the x-axis. This process forms a volume that can best be understood by visualizing the region turning over the horizontal axis, creating layered disks, resembling a pile of washers.
Real-world applications of this concept include finding volumes of objects like vases, bowls, or any similarly shaped items. This approach's beauty lies in its ability to simplify the visualization and calculation of complex three-dimensional structures by breaking them into manageable components.
The method of rotation influences the shape and size of the resulting solid. In the exercise, the area under the curve \( y = \sqrt{2x} \) between \( x = 0 \) and \( x = 2 \) is "swept" around the x-axis. This process forms a volume that can best be understood by visualizing the region turning over the horizontal axis, creating layered disks, resembling a pile of washers.
Real-world applications of this concept include finding volumes of objects like vases, bowls, or any similarly shaped items. This approach's beauty lies in its ability to simplify the visualization and calculation of complex three-dimensional structures by breaking them into manageable components.
Other exercises in this chapter
Problem 35
Approximate $$\int_{-2}^{2}\left(2+x^{2}\right) d x$$ using four equal subintervals and left endpoints.
View solution Problem 36
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{x^{3}}^{x^{4}} \ln \left(1+t^{2}\right) d t, x>0 $$
View solution Problem 36
Approximate $$\int_{-1}^{2} e^{-x} d x$$ using three equal subintervals and midpoints.
View solution Problem 37
Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2-x^{2}}^{x+x^{3}} \sin t d t $$
View solution