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Chapter 6: Problem 50

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

Expert verified
a. \( V = \frac{\pi}{3} b h^2 \); b. \( V = \frac{\pi}{3} h b^2 \)

Step by step solution

01

Define the region for revolution

The triangular region is formed by the vertices (0,0), (b,0), and (0,h). The lines forming the triangle are the x-axis, y-axis, and the line connecting (b,0) to (0,h), which has the equation y = -\(\frac{h}{b}\)x + h.
02

Set up the integral for revolution around the x-axis

For revolution around the x-axis, the triangular region defined in terms of the function y(x) = -\(\frac{h}{b}\)x + h is used. The volume is given by the integral \[ V = \pi \int_{0}^{b} (y^2) \, dx = \pi \int_{0}^{b} \left(-\frac{h}{b}x + h\right)^2 dx \]
03

Compute the integral for revolution around the x-axis

Expand the square inside the integral:\[ \left(-\frac{h}{b}x + h\right)^2 = \frac{h^2}{b^2}x^2 - \frac{2h^2}{b}x + h^2 \]Perform the integration:\[ V = \pi \left[ \frac{h^2}{b^2} \cdot \frac{x^3}{3} - \frac{2h^2}{b} \cdot \frac{x^2}{2} + h^2 x \right]_{0}^{b} \]Evaluate the integral:\[ V = \pi \left( \frac{h^2}{b^2} \cdot \frac{b^3}{3} - \frac{2h^2}{b} \cdot \frac{b^2}{2} + h^2 \cdot b \right) \]Simplify to get:\[ V = \frac{\pi}{3} b h^2 \]
04

Set up the integral for revolution around the y-axis

For revolution around the y-axis, express the function x(y) from the line:\[ x = -\frac{b}{h}y + b \]The volume is given by the integral:\[ V = \pi \int_{0}^{h} (x^2) \, dy = \pi \int_{0}^{h} \left(-\frac{b}{h}y + b\right)^2 dy \]
05

Compute the integral for revolution around the y-axis

Expand the square inside the integral:\[ \left(-\frac{b}{h}y + b\right)^2 = \frac{b^2}{h^2}y^2 - \frac{2b^2}{h}y + b^2 \]Perform the integration:\[ V = \pi \left[ \frac{b^2}{h^2} \cdot \frac{y^3}{3} - \frac{2b^2}{h} \cdot \frac{y^2}{2} + b^2 y \right]_{0}^{h} \]Evaluate the integral:\[ V = \pi \left( \frac{b^2}{h^2} \cdot \frac{h^3}{3} - \frac{2b^2}{h} \cdot \frac{h^2}{2} + b^2 h \right) \]Simplify to get:\[ V = \frac{\pi}{3} h b^2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral calculus
Integral calculus is a powerful branch of mathematics that allows us to compute things like areas under curves and volumes of solids. It works by summing infinitely small slices of a quantity to find totals. This process utilizes integrals, which are based on the concept of anti-differentiation.
The integral symbol \[ \int \] represents summation, and it's often seen alongside a function within a certain range, which tells us the bounds of the area or volume we are calculating.
  • An indefinite integral gives us a general form of a primitive function, essentially undoing differentiation.
  • A definite integral, on the other hand, has upper and lower bounds and provides a specific numerical result.
This capability is essential in many fields, from engineering to physics, where understanding changes and accumulations over time or space is vital.
Solid of revolution
The concept of a solid of revolution refers to a 3D shape created by rotating a 2D plane figure around an axis. Imagine spinning a straight line or a curve around an axis and "sweeping out" a volume.
To compute the volume of such a solid, we use integral calculus. We consider the small disks (or washers) that make up the structure as we slice across the axis.
  • Each slice, in the shape of a disk or washer, has a tiny volume that adds up to the total volume of the solid.
  • The thickness of these slices approaches zero, ensuring an exact total when summed through integration.
The mathematical representation requires setting up an integral that spans the bounds of the region being revolved. By solving this integral, one obtains the total volume of the solid.
Definite integral
A definite integral quantitatively measures the accumulation of a quantity over a given range. This integral type is written as \[ \int_{a}^{b} f(x) \, dx \], where \([a,b]\) specifies the interval. It essentially sums the small elements \((f(x) \, dx)\) of a function over a region.
To calculate the volume of a solid of revolution, like the one discussed using integration, definite integrals are utilized.

Step by Step Calculation

  • The process starts by identifying the bounds of integration based on the performance on the x or y axis.
  • The function being integrated represents the area or shape being revolved.
  • Finally, the integral simplifies to give the total volume or area within those bounds.
This approach is instrumental in finding total values in situations modeled by continuous functions, ensuring incompletely observable areas are accurately measured.

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Most popular questions from this chapter

The volume of a torus The disk \(x^{2}+y^{2} \leq a^{2}\) is revolved about the line \(x=b(b>a)\) to generate a solid shaped like a doughnut and called a torus. Find its volume. (Hint: \(\int_{-a}^{a} \sqrt{a^{2}-y^{2}} d y=\pi a^{2} / 2,\) since it is the area of a semicircle of radius \(a .)\)

A bag of sand originally weighing \(600 \mathrm{N}\) was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to \(6 \mathrm{m} .\) How much work was done lifting the sand this far? (Neglect the weight of the bag and lifting equipment.)

Find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region in the first quadrant bounded by \(x=y-y^{3}, x=1\) and \(y=1\) about a. the \(x\) -axis b. the \(y\) -axis c. the line \(x=1\) d. the line \(y=1\)

Find the volume of the torus generated by revolving the circle \((x-2)^{2}+y^{2}=1\) about the \(y\) -axis.

The square region with vertices \((0,2),(2,0),(4,2),\) and (2,4) is revolved about the \(x\) -axis to generate a solid. Find the volume and surface area of the solid.
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