Problem 50 - 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. a. Find the volume of the bowl. b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

Step-by-Step Solution

Answer

The volume of the bowl is \( 25\pi \). When the water is 4 units deep, the water level rises at \( \frac{3}{8\pi} \) units per second.

Step 1: Express x in Terms of y

The curve of the bowl shape is given by \( y = \frac{x^2}{2} \). We start by solving for \( x \) in terms of \( y \). By rearranging the equation, we get:\[x = \sqrt{2y}\]This will allow us to calculate the cross-sectional area of the bowl at any height \( y \).

Step 2: Set Up the Integral for Volume

We will use the method of disks to find the volume of the bowl. The formula for the volume of a solid of revolution around the y-axis is:\[V = \pi \int_{y=0}^{y=5} [r(y)]^2 \, dy\]where \( r(y) \) is the radius which is given by \( x \). From Step 1, we have \( r(y) = \sqrt{2y} \). Thus,\[V = \pi \int_{y=0}^{y=5} (\sqrt{2y})^2 \, dy = \pi \int_{y=0}^{y=5} 2y \, dy\]

Step 3: Calculate the Integral to Find the Volume

Integrate \( 2y \) with respect to \( y \):\[\int 2y \, dy = y^2\]Apply the limits of integration from 0 to 5:\[V = \pi [ y^2 ]_{0}^{5} = \pi [(5)^2 - (0)^2] = 25\pi\]Therefore, the volume of the bowl is \( 25\pi \, \text{cubic units} \).

Step 4: Set Up Equation for Related Rates

We know water is added at \( \frac{dV}{dt} = 3 \, \text{cubic units per second} \). We need to find \( \frac{dy}{dt} \) when \( y = 4 \). From Step 2, the volume of water as a function of height \( y \) is given by \( V(y) = \pi y^2 \). Differentiate with respect to \( t \):\[\frac{dV}{dt} = 2\pi y \frac{dy}{dt}\]Plug \( \frac{dV}{dt} = 3 \) and \( y = 4 \) into the equation:\[3 = 2\pi (4) \frac{dy}{dt}\]

Step 5: Solve for dy/dt

Solve \( 3 = 8\pi \frac{dy}{dt} \) to find the rate at which the water level rises:\[\frac{dy}{dt} = \frac{3}{8\pi}\]So, when the water is 4 units deep, the rate at which the water level is rising is \( \frac{3}{8\pi} \, \text{units per second} \).

Key Concepts

Integral Calculus

Integral calculus is fundamental in calculating the volume of complex shapes, such as the bowl in this problem. The process involves combining tiny slices across a region to derive quantities like area and volume. For solids of revolution, the integral allows us to sum an infinite number of infinitesimally small disks to find the final volume.
In this exercise, the function given is expressed in terms of another variable, with the shape generated by revolving this function around an axis. Through integration, we transform the shape's continuous nature into a calculable area by summing these slices over a set boundary.

• In Step 2 of the solution, the integral \( V = \pi \int_{y=0}^{y=5} 2y \, dy \) is set up for the computation of the bowl's volume.
• The integral represents a sum of all disk areas along the y-axis from 0 to 5.

This approach simplifies complex volume calculations and highlights the efficiency of integral calculus in handling continuous curves.

Related Rates

Related rates, a concept from calculus, deal with how different quantities change with respect to each other over time. In many practical scenarios, like filling a bowl with water, one needs to understand how the change in one variable affects another.
Here, the rate at which the water is added to the bowl is known, and we wish to find the rate of rise in water level. The exercises use derivatives to link variables like volume and water height, captured in \( \frac{dV}{dt} = 3 \, \text{cubic units per second} \).

• From Step 4, with the given volume equation \( V(y) = \pi y^2 \), differentiation helps relate the change in volume to the change in water height \( \frac{dy}{dt} \).

This method of solving related rates equations is invaluable in solving many real-world problems where the interaction of changing quantities is important.

Solids of Revolution

Solids of revolution involve creating 3D shapes by rotating a 2D area around an axis. The exercise at hand demonstrates this by spinning a curve about the y-axis to form a bowl. Understanding this concept allows students to visualize how dimensional transformation happens through revolution in mathematics.
By assembling slices or disks perpendicular to the axis of rotation, the area under the curve generates the volume when revolved. In our exercise:

• The function \( y = \frac{x^2}{2} \) gets revolved to craft the bowl.
• With cut sections of height \( dy \), each disk from Step 2 accumulates these cross-sectional areas into the total 3D shape stability.

Visualization and mathematical calculation unite to turn abstract areas into concrete volumetric forms.

Method of Disks

The method of disks is a technique used to find the volume of a solid of revolution. It involves slicing the solid into thin, circular disks perpendicular to the axis of rotation. The volume of each disk is calculated and then integrated along the axis to find the total volume.
In this problem, the disks are created by rotating the function \( y = \frac{x^2}{2} \) around the y-axis. The radius of each disk is given by \( x \), expressed as \( x = \sqrt{2y} \) from Step 1. The area of a typical disk is \( \pi r^2 \), and thus:

• From Step 2, the integral \( V = \pi \int_{y=0}^{y=5} (\sqrt{2y})^2 \, dy \) computes the volume.
• The simplicity and applicability of the disk method make it powerful for solving volumes in structured, rotational shapes.

Using the method of disks, one can tackle various problems involving volumes of solids with known functions, offering a systematic approach to complex spatial understanding.