Problem 50 - Calculus of a Single Variable

Question

Describing a Solid In Exercises 49 and \(50,\) the integral represents the volume of a solid. Describe the solid. $$ \pi \int_{2}^{4} y^{4} d y $$

Step-by-Step Solution

Answer

The solid described by the integral \(\Pi \int_{2}^{4} y^4 dy\) is a figure bounded by the planes y = 2 and y = 4. The height of the solid at any point on this axis is defined by the \( y^4\) function. Meanwhile, the shape in the \(x-z\) plane (perpendicular to y-axis) is circular, hence \(\Pi\) in the integral and volume. The total volume of the solid is proportional to 736\(\Pi\).

Step 1: Identify integral bounds

First, recognize the lower and upper bounds of the integral. In this case, they are 2 and 4 respectively. This means that the solid is bounded by the planes y=2 and y=4

Step 2: Interpret the integrand

The integrand function, \(y^4\), implies that the height of the solid at any given \(y\) value is proportional to \(y^4\). This yields a non-linear effect on the solids height.

Step 3: Evaluate the integral

In this step, evaluate the integral of \(y^4\) from 2 to 4. This gives the volume of the solid figure: \[\Pi \int_{2}^{4} y^4 dy = \Pi [\frac{1}{5}y^5]_{2}^{4}\]. This evaluates to \( \Pi (\frac{1}{5}\cdot4^5 - \frac{1}{5} \cdot 2^5 ) = 768 \Pi - 32 \Pi = 736 \Pi \]. This tells us the relative magnitude of the volume of the described solid.

Step 4: Describe the solid

Based on the given integral and its evaluated value, the solid can be described as: A shape bounded by the planes y = 2 and y = 4 along the y-axis. The height of the solid at any point along this axis is given by the function \( y^4\) and the volume is proportional to 736 \(\Pi\). The shape along \(x-z\) plane (perpendicular to y-axis) is circular, hence the \(\Pi\) in integral and volume.

Key Concepts

Definite Integrals

A definite integral represents the accumulation of quantities, like area under a curve or, in this case, the volume of a solid. The notation \(\int_{a}^{b} f(y) \, dy\) indicates that we are finding the accumulated value of \( f(y) \) as \( y \) moves from \( a \) to \( b \).

• In our integral, \( a = 2 \) and \( b = 4 \), so the solid is contained between these planes on the y-axis.
• The function in the integral, \( y^4 \), explains how the properties of the solid change as \( y \) varies.

The result of this integral, when multiplied by \( \pi \), gives the volume of the solid. This use of a definite integral highlights its power in measuring more than just area, extending into three-dimensional space.

Integration Techniques

Integration, especially of polynomial functions like \( y^4 \), involves applying rules that simplify the operation. The integral \( \int y^4 \, dy \) uses the power rule for integration:
\[ \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \]
Here, \( n = 4 \), so the integral becomes \( \frac{y^5}{5} \). This transformation was necessary before evaluating from 2 to 4.

• Substituting these bounds gives \( \left[ \frac{y^5}{5} \right]_{2}^{4} \), which means we measure the net accumulation by plugging in 4 and 2 into our expression, subtracting the lower from the upper.

Applying these techniques allows us to solve complex shape volumes in calculus by transforming them into more manageable calculations.

Solid Geometry

Understanding solid geometry is crucial for visualizing the solid described by the integral. The integral defines a solid of revolution, which means the solid is formed by rotating a curve around an axis.

• The planes \( y=2 \) and \( y=4 \) act as boundaries, limiting the solid's height along the y-axis.
• The function \( y^4 \) represents the radius of revolution, so as \( y \) goes from 2 to 4, this radius varies with \( y^4 \), leading to a cylindrical-like shape with a complex surface.

The geometry of these solids is not always intuitive, but understanding revolution helps in visualizing how the curves translate into physical forms.

Calculus Applications

Calculus allows us to explore and solve real-world problems by creating ideal models, such as solids of revolution. By representing 3D objects with integrals, we can calculate volumes and other properties efficiently.

• These calculations can be applied in physics for computing moments of inertia or in engineering design for material distribution.
• In biology, revolving integrals can model organ shapes or growth patterns of natural structures.

Using calculus in such applications showcases the range and practical utility of understanding integrals and integration techniques beyond academic exercises.