Question

An oil storage tank can be described as the volume generated by revolving the area bounded by \(y=\frac{16}{\sqrt{64+x^{2}}}, x=0, y=0, x=2\) about the \(x\) -axis. Find the volume of the tank (in cubic meters).

Short answer

The volume of the tank is approximately 0.3844 cubic meters.

Step-by-step solution

Set up the integral for the volume

To find the volume of the tank, we need to use the disk method to compute the volume of revolution about the x-axis. The formula for the volume is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). Here, \( f(x) = \frac{16}{\sqrt{64+x^{2}}} \), \( a = 0 \), and \( b = 2 \).

Calculate the integral

Calculate the definite integral using the formula \( V = \pi \int_{0}^{2} \left( \frac{16}{\sqrt{64+x^{2}}} \right)^2 \, dx \). Simplifying, we have:\[\pi \int_{0}^{2} \frac{256}{64 + x^{2}} \, dx = \pi \times 4 \int_{0}^{2} \frac{1}{64 + x^{2}} \, dx.\]

Recognize and apply the integral form

The integral \( \int \frac{1}{64 + x^{2}} \, dx \) is a standard form. It is similar to the integral \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left( \frac{x}{a} \right) + C \). Here, \( a^2 = 64 \), so \( a = 8 \). Substitute into the integral:\[4 \int_{0}^{2} \frac{1}{64 + x^{2}} \, dx = 4 \cdot \frac{1}{8} \left[ \arctan\left( \frac{x}{8} \right) \right]_{0}^{2}.\]

Evaluate the definite integral

Evaluate the expression:\[= \frac{4}{8} \left[ \arctan\left( \frac{2}{8} \right) - \arctan\left( \frac{0}{8} \right) \right] = \frac{1}{2} \left[ \arctan\left( \frac{1}{4} \right) - 0 \right] = \frac{1}{2} \arctan\left( \frac{1}{4} \right).\]

Calculate the final volume

Multiply the result from Step 4 by \( \pi \):\[V = \pi \cdot \frac{1}{2} \arctan\left( \frac{1}{4} \right).\]Note that \( \arctan\left( \frac{1}{4} \right) \approx 0.2449786631 \). Therefore, the volume is approximately:\[V \approx \frac{\pi}{2} \times 0.2449786631 \approx 0.3844 \text{ cubic meters}.\]

Key concepts

Disk Method

The Disk Method is an essential concept for finding the volume of a solid of revolution. It involves rotating a region bounded by a function around an axis, usually the x or y-axis. This technique is particularly helpful when the shape can be visualized as a series of disks stacked along the axis of rotation.

To apply the Disk Method:

• Identify the function and the axis of rotation. In this case, we have the function \(y = \frac{16}{\sqrt{64+x^{2}}}\) and we revolve around the x-axis.
• The radius of each disk is given by the function value \(f(x)\), which determines the distance from the axis to the curve.
• The volume of each disk is \(\pi[f(x)]^2\Delta x\), where \(\Delta x\) is the thickness of each disk.
• Integrate this disk volume expression from the lower boundary to the upper boundary of the region.

This process allows you to calculate the total volume of the solid formed by the rotation. The Disk Method is straightforward but requires careful setup of the integral based on the region being revolved.

Definite Integral

Understanding definite integrals is crucial when calculating physical quantities like area, displacement, or volume. A definite integral gives the accumulated value of a function, such as the volume of the solid body formed by revolving a curve around the x-axis.

For this exercise, the definite integral is expressed as:\[ V = \pi \int_{0}^{2} \left( \frac{16}{\sqrt{64+x^{2}}} \right)^2 \, dx \]This represents the sum of all the infinitely small disk volumes from \(x = 0\) to \(x = 2\).

Key points about definite integrals:

• They have both a lower and upper limit, indicating the region over which the accumulation occurs.
• The integral of a function squared over a particular interval gives a measure of how the shape contributes to the overall volume.
• Definite integrals can be solved using various techniques, including substitution and integration by parts, depending on the form of the function.

Mastery of these techniques helps in solving various real-world and theoretical problems.

Arctan Integration

Arctan integration is a technique used to solve integrals involving expressions of the form \(\frac{1}{a^2 + x^2}\). It is based on the inverse tangent function, \arctan\, which appears frequently in calculus problems involving trigonometric identities.

For the given exercise, once simplified, the integral needed is \[ \int \frac{1}{64 + x^{2}} \, dx \], which can be identified as \[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C \], where \(a\) is the constant \(8\) because \(a^2 = 64\).

Steps for Arctan Integration:

• Simplify the integral to match the standard form \(\frac{1}{a^2 + x^2}\).
• Identify the value of \(a\), then apply the formula.
• Evaluate the resulting expression at the given limits to find the definite integral.

This method helps to integrate functions that might otherwise appear daunting, making it a valuable tool in calculus.

Calculus Problem Solving

Calculus problem solving often involves breaking down complex functions and using integral calculus for finding solutions to real-world issues. In our case, we derived the volume of an oil tank by calculating a volume of revolution.

The process requires a strong grounding in various calculus techniques, including:

• Recognizing which methods to use: Whether you're using the Disk or the Shell method depends on the axis of revolution and the given function.
• Setting up the problem correctly: Ensure to set the correct limits of integration and function for accuracy.
• Simplifying integrations using known formulas and techniques, such as substitution or trigonometric identities.

This exercise combines several crucial calculus concepts: setting up an integral with the Disk Method, evaluating it using a definite integral, and applying Arctan integration. Each of these components requires attention to detail and a methodical approach to problem solving.