Volumes of a Region bounded by Two Curves

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SUMMARY

The discussion focuses on calculating the volumes of a region R bounded by the curves y = x² and y = x + 2, specifically when rotated about the x-axis and the line x = 2. For part (b), the volume using the disc method is correctly set up as the definite integral from -1 to 2 of π(x + 2)² dx, yielding a volume of 21π. For part (c), the cylindrical shell method is applied with the integral from -1 to 2 of 2πx(x + 2) dx, resulting in a volume of 12π. Participants confirm the use of the washer method for part (b) and the cylindrical method for part (c) as appropriate approaches.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with the disc method for volume calculation
  • Knowledge of the cylindrical shell method
  • Ability to sketch and identify points of intersection between curves
NEXT STEPS
  • Study the disc method for volume calculations in detail
  • Learn about the cylindrical shell method and its applications
  • Practice problems involving volumes of solids of revolution
  • Review calculus textbooks for examples related to rotating regions about axes
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Students studying calculus, particularly those focusing on volumes of solids of revolution, as well as educators seeking to enhance their teaching methods for these concepts.

doublehh06
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1. Let R be the region bounded by the curves y = x2 and y = x + 2.
(a) Sketch the region R and label the points of intersection between the two curves.
(b) Suppose we rotate R about the x-axis. Compute the volume of the resulting solid.
(c) What is the volume of the solid obtained by rotating the region R about the line x = 2?


Homework Equations



We need to know the equation for solving volume using a definite integral with the disc method and cylindrical method.

The Attempt at a Solution



I have part (a). For part (b), I set up the problem using the disc method which is the definite integral from [-1,2] of A(x) dx.

The set up looked like The definite integral from [-1,2] of pi(x+2)2dx.
After solving this, I got 21pi but am doubtful of my answer.

For part (c), I set up the problem using the cylindrical method which is the definite integral from [-1,2] of 2*pi*x (x+2)dx with radius x, circumference 2pi x and height x + 2. I got 12pi but am once again doubtful of my answer.

Am I on the right track? Help please?
 
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When you use the "washer" (disc) method, the volume of the washer is:

\pi(r_1^2-r_2^2)h

where the outer radius, r_1, is the line f(x)=x+2 and the inner radius, r_2, is g(x)=x^2 and the "height" of the infinitesimal disc would be dx since you're integrating with respect to x so the integrand would

\pi(f^2-g^2)dx

right?

Now for the cylindrical shell method, since you're going around the x-axis, can you see where that would require two integrals? The first "inner" integral would be the shells contributed by just the parabola up to the height where the line f=x+2 intersects it. Once you pass that, then the "height" of the shells would be determined by the distance between the line f=x+2 and the right side of the parabola or:

V=I_1+I_2=\int_0^1 (shell1)dy+\int_1^4 (shell2) dy

Maybe that's not very clear. How about just look in the calculus textbook on both of these methods. That would help too.
 
I have reviewed the topics in the book and feel I have an understanding; however, I have never seen problems like these before and can't relate them to examples in the book. I thought the disc/cylindrical methods may be appropriate, but I am not positive.

When you read the problem itself for the first time, how would you go about solving it?

Thanks.
 
doublehh06 said:
I have reviewed the topics in the book and feel I have an understanding; however, I have never seen problems like these before and can't relate them to examples in the book. I thought the disc/cylindrical methods may be appropriate, but I am not positive.

When you read the problem itself for the first time, how would you go about solving it?

Thanks.

Hi. I would solve it using the "washer" method. The volume of the infinitesimal washer is:

\pi(f^2-g^2)dx

right?

So then just integrate over the range -1 to 2.
 
That sounds right for part (B); however, would you use the same method for part (c)? Or would you try something else?
 
Hi Double. Sorry I'm not able to get to the web much these days. Maybe you've solved this already. Anyway, part C would best be solved using cylindrical shells and would only take one integral.
 

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