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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.2.59
Chapter 6, Problem 6.2.59

Find the area of the region described in the following exercises.


The region bounded by x=y(y−1) and y=x/3

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Step 1: Identify the curves that bound the region. The first curve is given as \( x = y(y - 1) \), which can be rewritten as \( x = y^2 - y \). The second curve is \( y = \frac{x}{3} \).
Step 2: Find the points of intersection between the two curves. To do this, substitute \( y = \frac{x}{3} \) into \( x = y^2 - y \). Replace \( y \) with \( \frac{x}{3} \) to get \( x = \left(\frac{x}{3}\right)^2 - \frac{x}{3} \). Solve this equation for \( x \) to determine the limits of integration.
Step 3: Set up the integral to calculate the area. The region is bounded vertically, so express the area as \( \int_{x_1}^{x_2} \left( \text{top curve} - \text{bottom curve} \right) dx \). Here, the top curve is \( y = \frac{x}{3} \) and the bottom curve is \( y = \sqrt{x + x^2} \) (after solving for \( y \) in terms of \( x \)).
Step 4: Determine the limits of integration \( x_1 \) and \( x_2 \) from the points of intersection found in Step 2. These values will serve as the bounds for the integral.
Step 5: Evaluate the integral \( \int_{x_1}^{x_2} \left( \frac{x}{3} - \sqrt{x + x^2} \right) dx \) to find the area of the region. Break the integral into manageable parts if necessary, and simplify before solving.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Defining the Region

To find the area of a region bounded by curves, it is essential to first understand the equations of the curves involved. In this case, the curves are given by x = y(y - 1) and y = x/3. Identifying the points of intersection between these curves will help determine the limits of integration for calculating the area.
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Integration

Integration is a fundamental concept in calculus used to calculate the area under curves. Once the region is defined by the intersection points, the area can be found by setting up an integral. Depending on the orientation of the curves, this may involve integrating with respect to x or y, which requires understanding how to express the functions appropriately.
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Area Between Curves

The area between two curves can be calculated by integrating the difference of the functions that define the upper and lower bounds of the region. In this case, once the curves are expressed in a suitable form, the area can be computed by integrating the difference between the two functions over the interval defined by their points of intersection.
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Related Practice
Textbook Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.


y=√sin x,y=1, and x=0; about the x-axis 

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Textbook Question

9–12. Consider the cylindrical tank in Example 4 that has a height of 10 m and a radius of 5 m. Recall that if the tank is full of water, then ∫₀¹⁰ 25 π ρg(15−y) dy equals the work required to pump all the water out of the tank, through an outflow pipe that is 15 m above the bottom of the tank. Revise this work integral for the following scenarios. (Do not evaluate the integrals.)


The work required to empty the top half of the tank

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Textbook Question

Leaky Bucket A 1-kg bucket resting on the ground contains 3 kg of water. How much work is required to raise the bucket vertically a distance of 10 m if water leaks out of the bucket at a constant rate of 1/5 kg/m? Assume the weight of the rope used to raise the bucket is negligible. (Hint: Use the definition of work, W = ∫a^bF(y) dy, where F is the variable force required to lift an object along a vertical line from y=a to y=b.)

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Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. 


{Use of Tech} y = √sin^−1x,y = √π/2, and x=0; about the x-axis

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Textbook Question

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.


R is bounded by y=1−x^3, the x-axis, and the y-axis.

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Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

x = 2y−4, for −3≤y≤4 (Use calculus, but check your work using geometry.)