Question on volume of a solid .... im confused..

X

xjinax07x

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In this problem..

Let region A be bounded by the y-axis, cosx, and (1/3)x+(1/2).

1) Find the area of region A.

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I know the answer to that is 0 to 0.786 [(cos(x)) - ((1/3)(x)+(1/2))]dx, but I don't understand why the answer is from 0 to 0.786 ... why wouldn't it be from -0.786 to 0.786 when there is a region in quadrant 2 too? I'm really confused. Please help! Thank you!!!
 
I did a quick graph of the problem, and looking the wording of the question, I see it as an ambiguous question.

There are 4 regions bounded by the 2 curves and the y-axis. What’s more, the problem doesn’t say to find the area to the x-axis or the y-axis. I'm not going to bother finding what region the book decided to choose because unless they are more specific, your answer is equally correct as the book since it fits the description of the bounds.

It sounds like you know what you’re doing, so I wouldn’t worry about it. If you can find the area of any of those regions you’re fine…just hope the question is more specific next time.
 
It seems clear to me except your title says volume and the problem says area. The area bounded by the y-axis, cos(x), and (1/3)x+(1/2) is sort of a triangle in the first quadrant. A "penant" flying off the y axis. I disagree with the domain. My TI-83 says cos(x)=x/3+.5 at x=.734 not .786 which is where the curves meet. The "triangle" has a base of .5 (y=.5 to 1) and a height of .734.
Your equation looks good and should come up with area = .213
 
Razor X said:
I did a quick graph of the problem, and looking the wording of the question, I see it as an ambiguous question.

There are 4 regions bounded by the 2 curves and the y-axis. What’s more, the problem doesn’t say to find the area to the x-axis or the y-axis. I'm not going to bother finding what region the book decided to choose because unless they are more specific, your answer is equally correct as the book since it fits the description of the bounds.

It sounds like you know what you’re doing, so I wouldn’t worry about it. If you can find the area of any of those regions you’re fine…just hope the question is more specific next time.
Click to expand...


Oh okay, so it should specify in more detail about whether you use one or more region? Also ... I graphed it and there are exactly TWO regions that are equal. after i get an answer for that question does that mean i would have to double the decimal answer? if there is another region that is equal in shape to that one?
 
Ok, well looking at it a second time (your right) there are actually only two regions that fit the definition (that is, two regions bounded by all three boundaries). However, like you said, there is a region in the first and second quadrant.

Now, it’s not both the regions together, because that combined region is not bounded by the y-axis, (the y-axis would be in the middle of the region, and the region needs the y-axis as a bound) so it’s one or the other. That’s the problem, which one? The answer is whatever one the book decided to choose. If they are equal in area, then either one works, just not both added together.
 
Razor X said:
Ok, well looking at it a second time (your right) there are actually only two regions that fit the definition (that is, two regions bounded by all three boundaries). However, like you said, there is a region in the first and second quadrant.

Now, it’s not both the regions together, because that combined region is not bounded by the y-axis, (the y-axis would be in the middle of the region, and the region needs the y-axis as a bound) so it’s one or the other. That’s the problem, which one? The answer is whatever one the book decided to choose. If they are equal in area, then either one works, just not both added together.
Click to expand...


Oh so whenever the problem says the region is bounded by the y axis .. it would have to b the region in the 1st quadrant right?
 
I agree with Razor that there is no mathmatical reason to choose one over the other, but I chose the first quadrant and would bet as high as a quarter that the careless author did the same. If he had tilted the line so that it didn't hit the cos in any other quadrant I could defend my choice better.
I have never understood why areas below the x-axis were negative but those to the left of the y-axis were not. I've always felt that areas in the second and forth should be negative and first and third positive.
 
Personally, on an exam, I'd do both and explain what I did and why. There is no substitute for being thorough.
 
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