Question

In Exercises 29-32, find the volume of the solid described. By integration, find the volume of the solid generated by revolving the triangular region with vertices \((0,0),(b, 0),\) \((0, h)\) about (a) the \(x\) -axis\(. (b) the \)y\( -axis\).

Short answer

The volume of the solid by revolving the triangular region about the x-axis and the y-axis is both \(\frac{1}{3} \pi b h\).

Step-by-step solution

Set up the equations

First, set up the equations for the triangle. The vertices of the triangle are \((0,0), (b, 0), (0, h)\). From these points, it can be inferred that the equation for the line representing the triangle is \(y = \frac{h}{b}x\).

Volume of solid revolving about the x-axis

To find the volume when the solid is revolved about the x-axis, use the formula \[ V = \pi \int_{a}^{b} y^2 dx\] . Plugging the equation \(y = \frac{h}{b}x\) and the limits of x-axis which is from 0 to b, then the formula becomes \[ V = \pi \int_{0}^{b} \left(\frac{h}{b}x\right)^2 dx\] . This can be solved by applying power rule.

Solve the integral

By expanding the equation above, we get \[ V = \pi \int_{0}^{b} \frac{h^2}{b^2}x^2 dx = \frac{\pi h^2}{b^2} \int_{0}^{b} x^2 dx\] . Applying the power rule on this yields to the volume of the cone revolving about the x-axis which gives \[ V = \frac{\pi h^2}{b^2} \left[\frac{x^3}{3}\right]_{0}^{b} = \frac{1}{3} \pi b h \] .

Volume of solid revolving about the y-axis

Next, compute for the volume when the solid is revolved about the y-axis. Similar to step 2, the formula is \[ V = \pi \int_{c}^{d} x^2 dy\] . However, expressing x as a function of y instead, we get \(x = \frac{b}{h}y\). The limits of y are from 0 to h. Hence, the equation to solve would be \[ V = \pi \int_{0}^{h} \left(\frac{b}{h}y\right)^2 dy\]

Solve the integral

Similar to step 3, expand the equation and solve the integral, we get \[ V = \pi \frac{b^2}{h^2} \int_{0}^{h} y^2 dy = \frac{\pi b^2}{h^2} \left[\frac{y^3}{3}\right]_{0}^{h} = \frac{1}{3} \pi b h \] . The volume of the cone revolving about the y-axis is similar to the volume of the cone revolving about the x-axis.

Key concepts

Volume by Integration

Calculating the volume of a solid by integration is a fundamental concept in calculus, particularly when dealing with irregularly shaped objects. This process utilizes the principle of slicing the solid into infinitesimally thin disks or washers, and then summing their volumes. The integral accumulates these small volumes over the range of the shape to find the total volume.

For the exercise with the triangular region, the problem is essentially looking to find the volume of a cone. By integrating across the relevant axis, either the x-axis or y-axis depending on the revolution, this gives a precise calculation of the volume that would otherwise be challenging to compute. It’s crucial to represent the radius of the disks or washers as a function of the variable of integration, which may require solving for one of the coordinates in terms of the other to properly set up the integral.

Rotational Volumes

Rotational volume refers to the volume of a 3D object that is created when a 2D shape is revolved around an axis. This is a powerful visualization tool because it helps conceptualize how a flat object can transform into a three-dimensional one through rotation.

In the case of the triangle in the exercise, when we revolve the shape about the x-axis or y-axis, we form a cone in either direction. By using the method of disks or washers, we can calculate the volume of these cones. It's crucial to ensure that limits of integration correctly correspond to the bounds of the shape along the axis of rotation. The iterative addition of each volume slice through the integral calculation gives the total volume of the solid.

Solids of Revolution

Solids of revolution are 3D solids obtained by rotating a 2D shape around an axis. The shape of the resulting solid depends on the profile of the original 2D shape and the axis of rotation. Common examples of solids of revolution include spheres, cylinders, and cones, just like the cone generated in the exercise when rotating a triangle around an axis.

Utilizing the methods of calculus to find the volume of solids of revolution is immensely helpful in various fields such as engineering, architecture, and even in manufacturing. It allows for the precise computation of complex volumes without the need to rely on geometrical formulas, which may be non-existent for more complicated shapes. When instructing students on this topic, it's beneficial to encourage them to visualize the rotation and comprehend the shape of the solid that forms, as it aids in setting up the relevant integral correctly.