Problem 49 - Single Variable Calculus

Question

Find the volume of the solid obtained by revolving the region enclosed by the curve \(y^{2}=\frac{1}{4}\left(2 x^{3}-x^{4}\right)\), where \(y \geq 0\), about the \(x\) -axis.

Step-by-Step Solution

Answer

The volume of the solid obtained by revolving the region enclosed by the curve \(y^2 = \frac{1}{4}(2x^3 - x^4)\), where \(y \geq 0\), about the x-axis is \(12\pi\).

Step 1: Set up the integral for the volume.

We need to find the volume generated by revolving the graph around the x-axis. We will use the "disk method". To do this, we can express the function as y and then get the integral for the volume. Since we are given \(y^2 = \frac{1}{4}(2x^3 - x^4)\), we can clear the square for y and obtain \(y = \sqrt{\frac{1}{4}(2x^3 - x^4)}\). Notice that this is only considering \(y \geq 0\), because we could have taken the negative square root as well, but it's not necessary here. Step 2: Find the bounds

Step 2: Find the proper bounds for the integral.

To find the proper bounds for the integral, we need to find where the graph intersects the x-axis. To do that, we will look for the x values when y=0: \(0 = \sqrt{\frac{1}{4}(2x^3 - x^4)}\) Squaring both sides, we get: \(0 = \frac{1}{4}(2x^3 - x^4)\) Multiplying both sides by 4, and then factorizing the expression, we have: \(0 = 2x^3 - x^4\) \(0 = x^3(2 - x)\) Thus, the intersection points are \(x = 0\) and \(x = 2\). These will be the bounds for our integral. Step 3: Set up the volume integral

Step 3: Set up the integral to find the volume with the disk method.

Using the disk method, the volume is given by the following integral: \(V = \pi \int_{a}^{b} (f(x))^2 dx\) We found the bounds to be \(x = 0\) and \(x = 2\). Thus, our integral becomes: \(V = \pi \int_{0}^{2} (\sqrt{\frac{1}{4}(2x^3 - x^4)})^2 dx\) Step 4: Solve the integral

Step 4: Solve the given integral.

Simplifying the square on the function within the integral, and perform the integral. \(V = \pi \int_{0}^{2} \frac{1}{4}(2x^3 - x^4) dx\) We can distribute the pi and the fraction inside the integral. We obtain: \(V = \frac{\pi}{4}\int_{0}^{2} (2x^3 - x^4) dx\) Now we can integrate: \(V = \frac{\pi}{4} \left[ \frac{1}{2}x^4 - \frac{1}{5}x^5 \right]_{0}^{2}\) Step 5: Evaluate the integral

Step 5: Evaluate the integral with the given bounds and find the answer.

We will plug in the bounds and subtract to find the volume. \(V = \frac{\pi}{4} \left[\frac{1}{2}(2)^4 - \frac{1}{5}(2)^5 - \left(\frac{1}{2}(0)^4 - \frac{1}{5}(0)^5\right) \right]\) Simplifying the terms, we have: \(V = \frac{\pi}{4}(16 - 64) = -12\pi\) However, we know volume cannot be negative, so the final answer is: \(V = 12\pi\) So the volume of the solid obtained by revolving the region enclosed by the curve \(y^2 = \frac{1}{4}(2x^3 - x^4)\), where \(y \geq 0\), about the x-axis is \(12\pi\).

Key Concepts

Disk Method

The disk method is a technique in integral calculus for finding the volume of a solid of revolution. Imagine slicing an object into thin disks perpendicular to a given axis. By summing up the volumes of these disks, you get the total volume of the solid. Here’s how it works:

• The volume of each infinitesimally thin disk is approximated as a cylinder: height equal to the infinitesimal width \(dx\) and radius \(f(x)\).
• The volume of one disk is \(\pi [f(x)]^2 dx\).

By integrating these disks along the axis from \(a\) to \(b\), we get the total volume of the object. Remember, this method is particularly helpful when the solid is symmetric about an axis. In our exercise, this method lets you calculate the volume by revolving a curve around the x-axis.

Integral Calculus

Integral calculus is a fundamental pillar of calculus dealing with integration. It focuses on the accumulation of quantities and the areas under and between curves. The specifics include:

• Integration as the inverse operation of differentiation.
• Finding antiderivatives of functions.
• Evaluating the area under a curve between two points.

The power of integral calculus lies in its ability to calculate things like volumes of solids, areas under complex curves, and even the work done by a force over a distance. In the context of our problem, integral calculus allows us to sum up the infinitesimal disk volumes to get the volume of a solid by using definite integrals.

Volume Integral

A volume integral is used to find the volume of a three-dimensional region. It is basically an application of multiple integrals to calculate volumes. In our specific scenario:

• We set up an integral that accounts for every small piece of volume across a defined region.
• The setup often follows using the disk or washer method, involving \(\pi [f(x)]^2\).

The process is to integrate over a given interval to sum all possible infinitesimal pieces, resulting in the total volume. The integral effectively handles the accumulation of infinitely small 3D space inside the object, greatly simplifying what would be an arduous geometric computation.

Definite Integrals

Definite integrals are used to compute a specific quantity, like area or volume, over a defined interval [a, b]. In a sense, they provide a numerical answer as opposed to a general formulation. Key aspects include:

• The integral’s limits \(a\) and \(b\) define the start and end points.
• The integration gives a net total of the function’s effect across that interval.

For our volume problem, definite integrals allow us to sum up all disk volumes between \(x = 0\) and \(x = 2\). This is instrumental in finding the exact total volume without making repetitive and complex calculations by hand. Definite integrals give a clear, precise, and manageable way to deal with accumulation functions in calculus.