Question

Sketch the region \(R\) bounded by the graphs of the given equations and show a typical horizontal slice. Find the volume of the solid generated by revolving \(R\) about the \(y\) -axis. \(x=2 \sqrt{y}, y=4, x=0\)

Short answer

The volume of the solid is \( 32\pi \) cubic units.

Step-by-step solution

Identify the Region

The region R is defined by the curves \( x = 2\sqrt{y} \), \( y = 4 \), and \( x = 0 \). This is a part of the plane bounded on the left by the line \( x = 0 \) (the y-axis), on the top by the line \( y = 4 \), and on the right by the curve \( x = 2\sqrt{y} \).

Determine Intersection Points

To find where the boundaries intersect, solve for \( y \) by setting \( x = 2\sqrt{y} = 0 \). This gives \( y = 0 \) as an endpoint. The figure is therefore bounded between \( y = 0 \) and \( y = 4 \). The intersection points at \( y = 4 \) occur at \( x = 0 \) and \( x = 2\sqrt{4} = 4 \).

Sketch the Bounded Region R

Draw the curves on a coordinate plane. The parabola \( x = 2\sqrt{y} \) opens to the right. It intersects the line \( y = 4 \) at \( x = 4 \). The region R is the area inside this parabola from \( y=0 \) to \( y=4 \) along with the line \( x=0 \).

Show a Horizontal Slice

A typical horizontal slice of width \( dy \) in R can be visualized at a height \( y \) as a finite line extending horizontally from \( x = 0 \) to \( x = 2\sqrt{y} \).

Set up the Volume Integral

The volume generated by revolving the region R around the y-axis can be calculated using the disk method. The radius of each disk at a height \( y \) is \( x = 2\sqrt{y} \) and its area is \( \pi(2\sqrt{y})^2 = 4\pi y \).

Evaluate the Volume Integral

Integrate the function for disk area from Step 5 with respect to \( y \) over the domain \( 0 \) to \( 4 \): \[ V = \int_0^4 4\pi y \, dy. \]Calculate this integral: \[ V = 4\pi \left[ \frac{y^2}{2} \right]_0^4 = 4\pi \left( \frac{16}{2} \right) = 4\pi \times 8 = 32\pi. \]

Conclusion: Volume of Solid

The volume of the solid generated by revolving the region \( R \) about the y-axis is \( 32\pi \) cubic units.

Key concepts

Disk Method

The Disk Method is a technique used in integral calculus to find the volume of a solid of revolution. It involves revolving a shape or region around a specified axis. In our case, we are revolving the region around the y-axis to form a solid. This method is best understood by imagining the solid as being made up of many small, thin disk-shaped slices. Each disk has a very small thickness of \( dy \) (or \( dx \), depending on the axis of revolution) and a radius that extends from the axis of rotation to the boundary of the region.
To calculate the volume of each infinitesimal disk, we use the formula for the area of a circle \( \pi r^2 \), where \( r \) is the radius of the disk. When the region is revolved around the y-axis, the radius at a given point \( y \) is the value of \( x \), which in this exercise is \( 2\sqrt{y} \). Therefore, the area of a disk at height \( y \) would be \( \pi (2\sqrt{y})^2 = 4\pi y \).
Finally, by integrating this area over the interval that defines the region, we can calculate the total volume of the solid.

Integral Calculus

Integral calculus is an essential branch of mathematics used to determine quantities like areas, volumes, and other related measures. In the context of finding the volume of solids of revolution, integration helps sum up infinitesimally small quantities over an interval to find a total.
In the given exercise, we used integration to sum the areas of an infinite number of disks that, together, form the solid generated by revolving the region \( R \) about the y-axis. The region is defined from \( y=0 \) to \( y=4 \), as determined from the intersection points and boundaries.
The specific integral set up to find the volume is given by \( V = \int_0^4 4\pi y \, dy \). Here, \( 4\pi y \) represents the area of a disk at a particular height \( y \). When this is integrated from 0 to 4, it determines the full volume of the solid. Performing the integral, we find that \( V = 32\pi \), indicating that the entire volume of the solid is \( 32\pi \) cubic units.

Parametric Equations

Parametric equations are an alternative way to express geometric shapes and curves. Rather than directly relating \( x \) and \( y \), parametric equations use a third variable, typically \( t \), to define both \( x \) and \( y \). This approach is very useful when dealing with complex curves or when needing clear control over a shape's motion and position.
In the context of the exercise, although not explicitly parametric, the curve \( x = 2\sqrt{y} \) can be thought of in a parametric form by introducing \( t \, (t = \sqrt{y}) \). Then, \( x(t) = 2t\) and \( y(t) = t^2 \). This way, for any given \( t \), you can calculate both the \( x \) and \( y \) coordinates. This format can simplify integration when evaluating complex trajectory or motion-related problems, but here it provides a clearer understanding of the curve's layout.

Sketching Regions

Sketching regions is an invaluable skill that assists with visualization in calculus. It can help you understand which area or volume you're dealing with, an essential step before setting up any integrations.
For the given problem, the sketched region is bounded by the curve \( x=2\sqrt{y} \), the horizontal line \( y=4 \), and the vertical line \( x=0 \). Visualizing this involves plotting each component:

• The curve \( x=2\sqrt{y} \) is a sideways-opening parabola.
• It is intersected horizontally by \( y=4 \), forming the upper boundary.
• The region is closed off on the left by the line \( x=0 \), which is the y-axis.

Once plotted, visualizing a horizontal slice helps reinforce the idea of disks being formed as the region is revolved around the y-axis. By clearly sketching the region, you understand the limits of integration, like from \( y = 0 \) to \( y = 4 \), and clarify how the rotation will form the solid.