Problem 52 - Thomas Calculus

Question

Volume of a bowl A bowl has a shape that can be generated by revolving the graph of \(y=x^{2} / 2\) between \(y=0\) and \(y=5\) about the \(y\) -axis. \begin{equation} \begin{array}{l}{\text { a. Find the volume of the bowl. }} \\ {\text { b. Related rates If we fill the bowl with water at a constant }} \\ {\text { rate of } 3 \text { cubic units per second, how fast will the water level }} \\\ {\text { in the bowl be rising when the water is 4 units deep? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. Volume of bowl: \( 25\pi \) cubic units. b. Water rises at \( \frac{3}{8\pi} \) units/sec for 4 unit depth.

Step 1: Set up the volume integral

First, identify the function that describes the edge of the bowl, which is given by the equation \( y = \frac{x^2}{2} \). To solve for \( x \), rearrange it to \( x = \sqrt{2y} \). The bounds for \( y \) are from 0 to 5 since these are the limits over which the function is rotated around the \( y \)-axis.

Step 2: Use the disk method

The volume \( V \) of the solid of revolution formed by rotating a curve around the \( y \)-axis can be calculated using the integral formula \( V = \pi \int_{a}^{b} [f(y)]^2 \, dy \). Here, \( f(y) = \sqrt{2y} \), so we can substitute \( R(y) = \sqrt{2y} \). Therefore, \([f(y)]^2 = 2y\). Plugging the function into the formula gives the integral \( V = \pi \int_{0}^{5} 2y \, dy \).

Step 3: Evaluate the volume integral

Calculate the integral from Step 2: \( V = \pi \int_{0}^{5} 2y \, dy = \pi \left[ y^2 \right]_0^5 = \pi (5^2 - 0^2) = \pi \times 25 = 25\pi \). So the volume of the bowl is \( 25\pi \) cubic units.

Step 4: Set up related rates problem

Now consider the situation where water is poured into the bowl at a rate of 3 cubic units per second, and we need to find how fast the water level \( h \) is rising when the water is 4 units deep. The volume \( V \) as a function of water height \( h \) is given by \( V = \pi \int_{0}^{h} 2y \, dy \).

Step 5: Differentiate volume with respect to time

Using the volume function, differentiate \( V = 2\pi \left[ y^2/2 \right]_0^h = \pi h^2 \) with respect to time \( t \). Applying the chain rule: \( \frac{dV}{dt} = 2\pi h \frac{dh}{dt} \). Given that \( \frac{dV}{dt} = 3 \), solve for \( \frac{dh}{dt} \): \( 3 = 2\pi h \frac{dh}{dt} \).

Step 6: Solve for \( \frac{dh}{dt} \)

Substitute \( h = 4 \) into the differentiated equation: \( 3 = 2\pi (4) \frac{dh}{dt} \). Simplify to find \( \frac{dh}{dt} \): \( \frac{dh}{dt} = \frac{3}{8\pi} \). Thus, the rate at which the water level is rising is \( \frac{3}{8\pi} \) units per second when the water is 4 units deep.

Key Concepts

Disk Method

The Disk Method is a useful technique in calculus to find the volume of a solid of revolution. This involves rotating a region around an axis to create a three-dimensional object. Specifically, when we revolve a function or shape around the y-axis, we often use the Disk Method.
Consider a bowl shape that is formed by revolving the function \( y = \frac{x^2}{2} \) around the y-axis. To apply the Disk Method, we first need the function in terms of \( y \) since the rotation is about the y-axis. We solve for \( x \) to get \( x = \sqrt{2y} \).
The general formula for the volume of the solid formed is:

• \( V = \pi \int [f(y)]^2 \, dy \)

This formula is derived by considering each cross-section of the solid as a disk with radius \( f(y) \). The cross-sectional area is \( \pi[f(y)]^2 \), and integrating these areas accumulates them into the total volume. For our example, \( f(y) = \sqrt{2y} \), giving us a radius squared of \( 2y \). The resulting volume integral from 0 to 5 is \( V = \pi \int_{0}^{5} 2y \, dy \), which simplifies to \( 25\pi \) cubic units. This tells us the volume inside the bowl.

Related Rates

Related rates problems deal with the relationship between different rates of change within a problem. In the context of the bowl problem, once we have the bowl filled with liquid, the rate at which the height of the liquid changes is related to the rate at which the liquid is being added to the bowl.
In this problem, water is being poured into the bowl at a steady rate of 3 cubic units per second. To find how fast the water level rises at a certain depth, note that the volume function with respect to the height \( h \) is \( V = \pi h^2 \). This describes the volume inside the bowl at any water height \( h \).
Differentiating both sides concerning time \( t \), using the chain rule gives:

• \( \frac{dV}{dt} = 2\pi h \frac{dh}{dt} \)

Here, \( \frac{dV}{dt} \) is given as 3 cubic units per second. To find \( \frac{dh}{dt} \), which tells us how fast the water level is rising, substitute \( h = 4 \):

• \( 3 = 2\pi (4) \frac{dh}{dt} \)
• Simplify to get \( \frac{dh}{dt} = \frac{3}{8\pi} \)

Thus, when the water depth is 4 units, the water level rises at approximately \( \frac{3}{8\pi} \) units per second.

Calculus Integration

Calculus integration is a powerful mathematical tool that allows us to find areas, volumes, and other quantities that can be seen as accumulations or sums of infinitesimal parts. In the exercise about the bowl, integration permits us to compute the volume of a solid formed by revolution.
To solve such problems, we often set up an integral that describes the entire volume. With our function \( f(y) = \sqrt{2y} \), integration becomes essential to calculate it over the specified bounds. In this case, from \( y = 0 \) to \( y = 5 \). The integral we used was:

• \( \pi \int_{0}^{5} 2y \, dy \)

Solving this integral involves finding the antiderivative of \( 2y \), which is \( y^2 \), and evaluating it at the limits of integration gives us the total volume: \( \pi[5^2 - 0^2] = 25\pi \).
Integration also plays a vital role in related rates problems where the rate of change of one quantity is determined via its relation to another through a derivative. By reversing this process (anti-differentiation), we can understand how rates of change interact. Together with differentiation, integration provides a complete picture to solve various calculus problems.