What is the volume of curve between the ordinate 0 to 4 around the curve x = y?

Question

Prepp Team · Accepted Answer

Understanding the Volume of Revolution Problem

The problem asks for the volume of the solid generated when the area under the curve \(x = y\) between the ordinates \(y = 0\) and \(y = 4\) is revolved around an axis. Although the axis of revolution isn't explicitly stated as the x-axis or y-axis, the function is given in the form \(x = f(y)\), and the limits are given for \(y\). This strongly suggests revolving around the y-axis. When revolving around the y-axis, the Disk or Washer method uses integration with respect to \(y\).

The curve \(x = y\) is a straight line passing through the origin. The region being revolved is the area between this line, the y-axis (\(x=0\)), and the horizontal lines \(y=0\) and \(y=4\). This forms a triangle with vertices at (0,0), (0,4), and (4,4). Revolving this triangular region around the y-axis generates a cone.

Applying the Disk Method for Volume Calculation

When revolving a region bounded by the curve \(x = g(y)\), the y-axis, and the lines \(y=c\) and \(y=d\) around the y-axis, the volume \(V\) can be calculated using the Disk Method. The formula is:

\[V = \int_{c}^{d} \pi [g(y)]^2 dy\]

In this problem:

The curve is \(x = y\), so \(g(y) = y\).
The ordinates (y-limits) are from \(y = 0\) to \(y = 4\). So, \(c = 0\) and \(d = 4\).

Substituting these values into the formula, we get:

\[V = \int_{0}^{4} \pi (y)^2 dy\] \[V = \pi \int_{0}^{4} y^2 dy\]

Calculating the Definite Integral to Find Volume

Now, we need to evaluate the definite integral:

First, find the indefinite integral of \(y^2\):

\[\int y^2 dy = \frac{y^{2+1}}{2+1} + C = \frac{y^3}{3} + C\]

Next, evaluate the definite integral using the limits from 0 to 4:

\[\int_{0}^{4} y^2 dy = \left[ \frac{y^3}{3} \right]_{0}^{4}\]

Apply the Fundamental Theorem of Calculus (evaluate at the upper limit and subtract the evaluation at the lower limit):

\[\left[ \frac{y^3}{3} \right]_{0}^{4} = \left( \frac{4^3}{3} \right) - \left( \frac{0^3}{3} \right)\] \[= \frac{64}{3} - 0\] \[= \frac{64}{3}\]

Now, multiply by \(\pi\) to get the volume \(V\):

\[V = \pi \times \frac{64}{3}\] \[V = \frac{64\pi}{3}\]

Comparing the Calculated Volume with Options

We found the exact volume is \(\frac{64\pi}{3}\). Let's calculate the decimal approximation for \(\frac{64}{3}\):

\[\frac{64}{3} \approx 21.333...\]

So, the volume is approximately \(21.33 \pi\).

Let's look at the given options:

Option	Value	Comparison with \( \frac{64\pi}{3} \approx 21.33\pi \)
1	\(31.33\)	Does not match. This value does not include \(\pi\).
2	\(21 \pi\)	Close, but \(64/3\) is approximately 21.33, not exactly 21.
3	\(21.33 \pi\)	Matches the calculated approximate value.
4	\(31 \pi\)	Does not match.

The calculated volume \(\frac{64\pi}{3}\) is approximately \(21.33 \pi\), which matches Option 3.

Revision Table: Key Concepts in Volume of Revolution

Concept	Description	Formula (around y-axis)
Volume of Revolution	The volume of a solid formed by rotating a plane region around an axis.	-
Disk Method	Used when the solid has no hole (solid of revolution touches the axis of revolution) and the cross-sections perpendicular to the axis are disks. Integrate \(\pi \times (\text{radius})^2\).	\(V = \int_{c}^{d} \pi [g(y)]^2 dy\) (for rotation around y-axis, \(x=g(y)\))
Limits of Integration (Ordinates)	The y-values that define the boundaries of the region being revolved. Here, from y=0 to y=4.	\(c\) and \(d\) in the integral limits.
Radius \(R(y)\)	The distance from the axis of revolution (y-axis) to the curve \(x=g(y)\). Here, \(R(y) = x = y\).	\(g(y)\) in the formula.

Additional Information: Volume of Revolution Methods

Besides the Disk Method, another common technique for finding the volume of revolution is the Washer Method and the Shell Method.

Washer Method: This is used when the solid has a hole in the middle, meaning the region being revolved does not touch the axis of revolution throughout the entire interval. It involves subtracting the volume of the inner hole from the total volume of the outer shape. The formula for rotation around the y-axis is \(V = \int_{c}^{d} \pi ([R_{out}(y)]^2 - [R_{in}(y)]^2) dy\). In our case, the region touches the y-axis, so the inner radius is 0, and the Washer Method simplifies to the Disk Method.
Shell Method: This method calculates volume by summing the volumes of cylindrical shells. For rotation around the y-axis, the formula is \(V = \int_{a}^{b} 2\pi x h(x) dx\), where \(h(x)\) is the height of the shell at \(x\). For our problem \(x=y\), rotating around the y-axis could also be done using shells, integrating with respect to \(x\). The curve is \(y=x\), limits from x=0 to x=4. The height of a shell at \(x\) is \(y=x\). So \(V = \int_{0}^{4} 2\pi x (x) dx = 2\pi \int_{0}^{4} x^2 dx = 2\pi \left[ \frac{x^3}{3} \right]_{0}^{4} = 2\pi \left( \frac{64}{3} \right) = \frac{128\pi}{3}\). Wait, this is double the result from the disk method. This indicates that revolving the *region* bounded by x=y, x=0, y=0, y=4 around the y-axis using shells would be \(V = \int_{0}^{4} 2\pi x (4-x) dx\) for the triangle from (0,0) to (4,4) to (0,4), which generates a cone with the top cut off, plus a cylinder. The Disk method approach for \(x=y\) from y=0 to y=4 around the y-axis is generating a simple cone with height 4 and radius 4, matching our earlier calculation. The Shell Method formulation must be carefully applied based on the region and axis. The disk method is generally more straightforward when revolving \(x=g(y)\) around the y-axis.

Understanding which method to use depends on the shape of the region, the axis of revolution, and whether the function is easier to express as \(y=f(x)\) or \(x=g(y)\).