evaluating double integrals to find area


by footboot
Tags: double, evaluating, integrals
footboot
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#1
Nov20-09, 01:23 PM
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1. The problem statement, all variables and given/known data

Find the area of the region R bounded by the parabola y = 2x2−2
and the line y = 2x + 2 by sketching R, and evaluating the area
integral

A = (doubleintegral) R dxdy.

3. The attempt at a solution

i found the points of intersection which were -1 and 2 and sketched the region. but im really stuck on evaluating it. If it was dydx i would have no problem but since its dydx i keep ending up with x's in my answer!
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rock.freak667
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Nov20-09, 03:02 PM
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What did you put as the limits for the integrals?
tiny-tim
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Nov20-09, 03:37 PM
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Hi footboot! Welcome to PF!

(have an integral: ∫ and try using the X2 tag just above the Reply box )
Show us exactly what you've done.

footboot
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#4
Nov20-09, 04:59 PM
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evaluating double integrals to find area


im sorry its quite hard ot get used to writing maths on the internet!
The limits i used were 2 and -1 on the left integral and 2x+2 and 2x2 -2 on the right integral.
I integrated with respect to x which just gave x, then filled in the values and got
(2x+2)-(2x2-2) which gave -2x2 +2x +4
I then integrated with respect to y and filled in 2 and -1 for y but all the x's from the first parrt are still there so I get an answer of x2+x+10. Which is obviously not right as I presume the answer should have no variables in it?
tiny-tim
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#5
Nov20-09, 05:12 PM
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Hi footboot!
Quote Quote by footboot View Post
The limits i used were 2 and -1 on the left integral and 2x+2 and 2x2 -2 on the right integral.
I integrated with respect to x which just gave x, then filled in the values and got
(2x+2)-(2x2-2) which gave -2x2 +2x +4
I then integrated with respect to y and filled in 2 and -1 for y but all the x's from the first parrt are still there so I get an answer of x2+x+10. Which is obviously not right as I presume the answer should have no variables in it?
oh i see now

you've integrated in the wrong order!

integrate wrt y first, that gives you a function of x, then integrate wrt x and that gives you a number.

(alternatively, if you integrate wrt x first, you must do it between limits which are a function of y)
footboot
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#6
Nov20-09, 05:16 PM
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In the question it says dxdy so doesnt this mean i have to integrate wrt to x first? And if i do have to do it that way how would I find the limits as a function of y?
Thanks for your help!
tiny-tim
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#7
Nov20-09, 05:26 PM
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Quote Quote by footboot View Post
In the question it says dxdy so doesnt this mean i have to integrate wrt to x first? And if i do have to do it that way how would I find the limits as a function of y?
Without seeing the exact question, it's difficult to say.

dxdy = dydx, and the order of integration doesn't matter, so you can do it either way.

However, of course, if the question insists on integrating wrt x first, then it's checking whether you know how to get the limits (which you don't, yet! ), and of course you must do it that way.

So, what are the limits for x (they are functions of y) if you integrate wrt x first?

(in other words, for a fixed value of y, what are the limits of x?)
footboot
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#8
Nov20-09, 05:47 PM
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In lectures we have been doing examples of functions where dxdy is not equal to dydx so thats why i thought i couldnt do it here. I did integrate wrt y first anyway and came out with an answer of 9 which looks about right from the graph iv drawn.
By changing the limits for x did you mean change the original functions for the parabola and line? because I did that first and i came out with quite complicated terms for the limits! So im really hoping 9 is right
Thanks for all your help!
tiny-tim
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Nov20-09, 05:54 PM
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Quote Quote by footboot View Post
I did integrate wrt y first anyway and came out with an answer of 9 which looks about right from the graph iv drawn.
By changing the limits for x did you mean change the original functions for the parabola and line? because I did that first and i came out with quite complicated terms for the limits! So im really hoping 9 is right
Unless you show what you've done, I can't say whether it's right.
footboot
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#10
Nov20-09, 06:02 PM
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I integrated wrt to y and got y, filled in 2x+2 and 2x2-2 into y and got -2x2 +2x +4. Then integrated with respect to x and got -2x3/3 + x2 + 4x. filled in 2 and -1 to this function and ended up with a value of 9. Which is hopefully right as iv been spending hours on this question! Also I was wondering I did a different
integration area question earlier and got a negative answer even though the function was above the x axis would this indicate that I went wrong somewhere in the question?
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Nov20-09, 06:08 PM
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Quote Quote by footboot View Post
I integrated wrt to y and got y, filled in 2x+2 and 2x2-2 into y and got -2x2 +2x +4. Then integrated with respect to x and got -2x3/3 + x2 + 4x. filled in 2 and -1 to this function and ended up with a value of 9.
Yes, that looks ok.

(if you got a negative area in another problem, then yes, that's obviously wrong

I'd guess you got the two limits the wrong way round )


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