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3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x. I get 4653pi/105 after help from Mr. Reiny, but thi...Asked by Hannah
3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.
I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you
I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you
Answers
Answered by
Steve
The curves intersect st (-1,0) and (0,0) so the area is
∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
= 1/4 x^4 - x^3 - x^2 [-1,0]
= -1/4
Was there some rotation involved?
∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
= 1/4 x^4 - x^3 - x^2 [-1,0]
= -1/4
Was there some rotation involved?
Answered by
Hannah
No there wasn't. This is exactly the problem our substitute gave us, and I was having a hard time figuring out why that was so.
Do they not intersect at (2, 0) as well?
I was told this answer is incorrect as well (-1/4). I'm perplexed. :S
Do they not intersect at (2, 0) as well?
I was told this answer is incorrect as well (-1/4). I'm perplexed. :S
Answered by
Steve
Don't think so.
f(2) = 4
g(2) = 12
If the answer involves pi, there has to be some rotating involved. That answer must be for some other problem.
Check with your instructor. Something's amiss here.
f(2) = 4
g(2) = 12
If the answer involves pi, there has to be some rotating involved. That answer must be for some other problem.
Check with your instructor. Something's amiss here.
Answered by
Steve
I took a look at Reiny's solutions. He does assume you are revolving the region, so that's bogus.
I see I made a typo. The curves do also intersect at (2,12)
If you want algebraic (signed) area,
∫[-1,2] (x^3+x^2) - (2x^2+2x) dx
= -9/4
If you want total unsigned area, then you need
∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
+ ∫[0,2] (2x^2+2x)-(x^3+x^2) dx
= 5/12 + 8/3
= 37/12
So, if you got 5/12, it looks like you only figured the left-hand side of the area.
I see I made a typo. The curves do also intersect at (2,12)
If you want algebraic (signed) area,
∫[-1,2] (x^3+x^2) - (2x^2+2x) dx
= -9/4
If you want total unsigned area, then you need
∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
+ ∫[0,2] (2x^2+2x)-(x^3+x^2) dx
= 5/12 + 8/3
= 37/12
So, if you got 5/12, it looks like you only figured the left-hand side of the area.
Answered by
Hannah
I absolutely understand this now. Thank you so much. I really couldn't have done it without your help Steve.
Answered by
Steve
Glad to help. Gotta watch those typos, especially the copy/paste ones!
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