Question

Find the volume of the solid formed when the area under \(y=x^{2}\) between \(x=1\) and \(x=2\) is rotated about the \(x\) axis.

Short answer

Answer: The volume of the solid is \(\frac{31\pi}{5}\).

Step-by-step solution

Identify the function and the interval

The function is given as \(y=x^2\). The interval being considered is between \(x=1\) and \(x=2\).

Set up the integral for the disk method

Using the disk method, the volume \(V\) of the solid can be found by integrating the cross-sectional area of the infinitesimally small cylindrical disks over the given interval. The area of each disk is given by the formula \(A(x) = \pi [y(x)]^2\), where \(y(x)\) is the height of the disk at a given value of \(x\). In this case, \(y(x) = x^2\), so the area formula becomes \(A(x) = \pi (x^2)^2 = \pi x^4\). To find the volume, we need to integrate \(A(x)\) over the interval \([1,2]\): $$V = \int_{1}^{2} \pi x^4 dx$$

Integrate the function

Now we need to find the antiderivative of the function \(\pi x^4\) with respect to \(x\). Using the power rule, we get: $$\int \pi x^4 dx = \frac{\pi x^5}{5} + C$$ However, we're only interested in the definite integral on the interval \([1, 2]\).

Evaluate the definite integral

Using the Fundamental Theorem of Calculus, we can evaluate the definite integral by subtracting the value of the antiderivative at the lower bound from the value at the upper bound: $$V = \left[\frac{\pi x^5}{5}\right]_{1}^{2} = \frac{\pi (2)^5}{5} - \frac{\pi (1)^5}{5} = \frac{32\pi}{5} - \frac{\pi}{5} = \frac{31\pi}{5}$$

State the final answer

The volume of the solid formed when the area under the curve \(y=x^2\) between \(x=1\) and \(x=2\) is rotated about the \(x\)-axis is: $$V = \frac{31\pi}{5}$$

Key concepts

Disk Method

The disk method is a technique for finding the volume of a solid of revolution. Imagine the area under a curve is being rotated around an axis, like clay on a potter's wheel. When it spins, it forms a 3D shape. The disk method helps us calculate the volume of that shape by slicing it into thin disks or cylinders.
To find the volume, we need the area of each disk. The formula for the area at a point where the solid touches the curve is given by

• \(A(x) = \pi [y(x)]^2\)

In our exercise, the curve is \(y = x^2\), so the area of each disk is \(\pi x^4\). The volume is then the sum of these areas as we integrate over the interval from \(x = 1\) to \(x = 2\).
This is why the formula becomes:

• \[V = \int_{1}^{2} \pi x^4 dx\]

Definite Integral

A definite integral allows us to find the total volume under a curve between two points. It uses limits to determine the sum of infinitesimal areas between these points. In our problem, we are tasked with integrating the function \(\pi x^4\) from \(x = 1\) to \(x = 2\).
Definite integrals are denoted by:

• \[ \int_{a}^{b} f(x) \, dx \]

where \(a\) and \(b\) define the interval, and \(f(x)\) is the function being integrated. This helps us find exact values, which means calculating the exact total area represented between \(x = 1\) and \(x = 2\).
The result of this integral gives us the volume of the solid.

Power Rule

The power rule is a fundamental concept in calculus for finding antiderivatives. It's a shortcut that helps us integrate functions more easily. If you have a function \(x^n\), applying the power rule gives:

• \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

where \(C\) is the constant of integration (not needed for definite integrals).
In our problem, \(x^4\) becomes \(\frac{x^5}{5}\) when we apply the power rule. We multiply by \(\pi\) because it's part of the original function \(\pi x^4\). This makes it:

• \[ \int \pi x^4 \, dx = \frac{\pi x^5}{5} \]

This antiderivative is then used to evaluate the definite integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing how they are inverses. It has two main parts, but the one we need helps evaluate definite integrals.
When we have the antiderivative of a function, the theorem tells us that:

• \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

where \(F(x)\) is the antiderivative of \(f(x)\). In other words, we find \(F(x)\) and plug in our limits, then subtract.
In our exercise, \(F(x) = \frac{\pi x^5}{5}\). We evaluate this at the points:

• \(x = 2\) and \(x = 1\)

Finally, subtract: \(\frac{32\pi}{5} - \frac{\pi}{5} = \frac{31\pi}{5}\). This gives us the calculated volume.