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John_henderson
 3 years ago
V= ∫ ∫ 4dydx + ∫ ∫ 4dydx
For the first set of integrals the outer limits are from 1 to 0. The inner limits are from X^2 to 0.
For the second set of integrals the outer limits are from 2 to 1. The inner limits are from (X2)^2 to 0.
Sketch the limits of integration, and the shape of the prism, Change the order of limits and solve the volume of the prism.
I need help on changing the order of integration and sketching the prism. Please Help.
John_henderson
 3 years ago
V= ∫ ∫ 4dydx + ∫ ∫ 4dydx For the first set of integrals the outer limits are from 1 to 0. The inner limits are from X^2 to 0. For the second set of integrals the outer limits are from 2 to 1. The inner limits are from (X2)^2 to 0. Sketch the limits of integration, and the shape of the prism, Change the order of limits and solve the volume of the prism. I need help on changing the order of integration and sketching the prism. Please Help.

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John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0\[Volume=\int\limits_{0}^{1}\int\limits_{0}^{x^2} 4dydx + \int\limits_{1}^{2}\int\limits_{0}^{(x2)^2}4dydx\]

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0I need the order to be as dxdy? and sketch the prism ? i guess the vol= pi*y^2

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0the way the problem is given you are not doing a surface revolution, so we can supposed that 4 represents the height, so it's really just about changing the integration order have you got a sketch yet?

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Oh yes the lengtth is 4cm sorry i missed that out!

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0let me skecth it...

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354460197419:dw

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Some thing like that for the first area?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0you should really draw the two graphs on top of one another to get a feel for the situationdw:1354460240069:dw

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0so you can see why they call it a "prism" so, where is the intersection point?

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Thats what i did , first but the area of region was o the left.. anyway i guess the limits are going to be x=rooty to 0 , 1 to 0 and x=rooty+2 to 2,1 ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0almost, but look at a strip for x's bounds...dw:1354460805961:dwsince x is bounded by the left half of the second graph we need the negative square root now this can be done with just one double integral instead of two; what are the new bounds on y?

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0so do you equate the limits in order to form one double integral? , so i think y= is going to be sqrt(x^2) to 2sqrt(x2)^2))

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0why are you taking the square root of x^2 ? just look at the bounds x is bounded below by sqrt(y) and above by 2sqrt(y), so those are the limits for dx to find the limits on y, all you need is the intersection point of the two graphs. what is the range of y values in the region D ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0what is this point?dw:1354461347743:dw

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.01 to 0 root2 to 0?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0y cannot be negative, and you are giving two sets of bounds. y only has one set of bounds; again find the point of intersection! btw, you should also always say "upper" to "lower" as in "0 to 1", not "1 to 0" unless you specifically intend to integrate negatively

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I mean lower to upper*

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry , 0 to 1, . 0 to 3+root2

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0im not sure as there are 2 seperate coordinates ?which do i sub in?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0okay, I still don't see why you have two sets of bounds, and why they are both numerical. the inner bounds on dx are in terms of y, and we have already found them the outer bounds should be the numerical limits for y in the region

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354461728579:dwso x is bound as follows\[\sqrt y\le x\le2\sqrt y\]and y is simply bound between the xaxis and the point of intersection, at which y=1\[0\le y\le1\]so that's your new integration, all one integral with those bounds.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0one double integral of course*

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0oh right so your dont have to compute the values of x=2,1 why is that?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0because with the first way we were doing it, we had two different upper bounds for y as we went along x; at first the above for y was bound by x^2, then at x=1 it changes to being bound above by (x2)^2 this change in our upper bounds forced us to use two integrals when integrating along x last after changing the order, x is bound above the whole time by 2sqrt(y) and below the whole time by sqrt(y), hence only one integral is necessary.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0here is the first integraldw:1354462159166:dw

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354462244107:dwso going along x last, this is why we needed two integrals....

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0when integrating along y last we have no such division point, the upper and lower bound on x stay the same

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0right i see so the values of x=1 , 2 are subed into (x2)^2 to get 0 to1..

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0you could look at it that way, or you could just look at the pic (you may have to run your head sideways to see the shape along y) and see that y is always somewhere between 0 and 1, since the intersection point is (1,1)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0your way will not always work actually, those values are coincidental to some extent why not sub into the other function, x^2, by your logic?

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0I have finally done got the ans and i got 8/3 i guess i need to multiply this by pi(8/3)^2 =v so get the volume?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354462650777:dwto me, the best way is to say "what is this range, here?" and visually you can get it

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0no you don't need to multiply by anything if 4 is the function for the height. This is not a volume of revolution! let me try the integral myself...

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0yes is can clearly see its from o to 1.. its much easir sketching it !

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I always sketch it anyway, I get the same as you; 8/3 since we are NOT revolving this shape around any axis, this is not going to be one of those cal1type tricks where we need to use the formula for a circle (in which case we would have needed to have that in the integral, remember?). Actually, the whole point of multivariable calc is that we don't need the revolution trick at all, that is a beginners method only. This shape is not circular; it has an base area D (which we find by integrating dxdy in the graph) and a height of 4 so we just do base times height here, i.e. 4 times the double integral

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0I have got a answer for the voume which is 64/9 pi, does that sound about right. I integrated with respect to dxdy?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0but the 4 in the problem as you are given it is most likely representing the height, so 8/3 should be the final answer.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0no, it sounds very wrong, read what I wrote: this is NOT A VOLUME OF REVOLUTION

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Damn seriously i thought it was the volume of revolution? , right i undertand now basically 8/3 is the base and 4=h , well if you multiply them together dont you get area?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354463282138:dwthis is what the shape sort of looks like. pi should not be involved as this is not circular

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0whether 8/3 is the final answer or not depends on what the 4 represents in your initial problem

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0I finally get it lol the inregral is the area = 8/3 and the h=4cm so v=a*h v=32/3 cm^3

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0the area of the base D is just\[A=\int_0^1\int_{\sqrt y}^{2\sqrt y}dxdy\]so I think the 4 in the integral represents height. If not, then why is it there?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0no, listen to what I am saying, you need to look more closely at what your problem is telling you how exactly is the question worded?

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0The volue of an unusally shaped prism of length 4cm can be evaluated by using the sum of two double integrals as show above. A. Sketch the region bounded by the limts of integration of the ingreals to illustrate the shape of the crosssection of the prism. B. Change the order of integration and adjust the limits appropriately for the integrals in a and express the volume as a single double interal. hence evaluate the integral and obtain the volume of the prism.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0okay, so the initial two integrals you are given represent the *volume*, not just the base area "The volume of an unusually shaped prism of length 4cm can be evaluated by using the sum of two double integrals as show above." that tells us that the 4 in the integral must be the height (once you get to triple integrals you will see that this is the same as integrating in the z from 0 to 4)

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Right i see i have never done triple integral before, i dont know anything about that. Anyway i guess the overall answer 32/3 then?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0no, you are multiplying by 4 again, but the initial set of two integrals you are given, we are told, represent VOLUME not AREA since the area is given by just\[A=\int_0^1\int_{\sqrt y}^{2\sqrt y}dxdy\] (notice the area does not have 4 in the integrand) the volume must be\[Ah=A(4)=\int_0^1\int_{\sqrt y}^{2\sqrt y}4dxdy=8/3\]and that's it

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.02/3 divide it by 4

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0what is is purpose of 4 ? so i guess you leave it as 8/3 =v ,v=AH, well in the question it says volume so i think it remains the same?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0yes, that is my whole point

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Oh i see that means the area is 2/3 which is a H=4 , which therefore gives 8/3 , Wow this is real trick question huh? Its so easy to make mistakes!

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0I guess you really need to read the question Lol..!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0the integral\[A=\iint\limits_D dA=\int_a^b\int_{f(x)}^{g(x)}dydx\](with NO INTEGRAND) is the area, so the 4 MUST be the height, so 8/3 is the final answer. yes, I agree that the wording of the question is meant to throw you off, but I'm quite certain that they mean what you said; the area is 2/3 and height is 4

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for your time, i really understand now. I was wondering if your would like to work though another question with me on fourier transform?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know Fourier transform. I know Fourier series and Laplace transform, did you mean one of those by chance.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0and you're welcome :)

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0Yes is mean fourier series sorry.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0okay, it's been a reeeeeeeally long time for me with Foiurier series, but I'll have a look.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0you should really post the problem separately, as a new thread

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0A periodic funtion F(t) is defined by f(t)=t^2, over pi<t<pi A1. state the period of f(t). 2. sketch f(t) over the domain 3pi<t<3pi B. state with reasons, wheter f(t0 is odd , even or neither odd or even. C. dertermine the first six non zero terms of the fourier series represting f(t).

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think I know how to do this though...

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0For be its meant to be f(t) only not f(t0

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0it is fourier series not transform.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0It's still been too long for me; my notes are too foreign looking for me to recover how to do this right now. You should post this separately anyway.

John_henderson
 3 years ago
Best ResponseYou've already chosen the best response.0i got 2 for the period, the sketch is 3times y= pulus or minus y=x^2 the function is even . but i need help onlt of the series please.
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