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what is double integral function?
how to solve double integral function?
 one year ago
 one year ago
what is double integral function? how to solve double integral function?
 one year ago
 one year ago

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ViPhyBest ResponseYou've already chosen the best response.0
Double Integral: http://mathworld.wolfram.com/FubiniTheorem.html You solve in the order of the integral, if dy comes first, you solve as simple integral considering x as constant, after that, solve the integral for dx.
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
is the value of dy similar to the value of dx?
 one year ago

yunusBest ResponseYou've already chosen the best response.0
yes they are similar. when you solve the integral according to dy all x values are constant for you. It doesnt matter which one you solve first, dx or dy.
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
ahh..ok tnx..can you teach me how to solve this?i have no background in solving this,, \[\int\limits_{0}^{2}\int\limits_{0}^{x ^{2}/2} x/\sqrt{1 + x^2 + y^2} dydx\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
if the limits for both the variables were constant, then we can integrate any variable first., but here since the limits of inner integral is not constant, we need to evaluate the inner integral first, limits are y=0 to y=x^2/2 so, since you are integrating w.r.t 'y', treat x as constant, so you have, \(\int\limits_{0}^{2}x[\int\limits_{0}^{x ^{2}/2} 1/\sqrt{1 + x^2 + y^2} dy]dx\) now the inner integral is od the form, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy\) do you know how to integrate this ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
i just took 1+x^2 = constant = a^2 because we have a standard integration formula for, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
then we need to resubstitute back after integration, a^2= 1+x^2 before integrating w.r.t x
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
*after integration w.r.t y,
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
in this we are integrating y and we assume as x as constant?
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
what do you mean by w.r.t y, ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
yes, true, so we'll get answer in 'x' after first integrating w.r.t y (with respect to y)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
ok, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy= (1/a)\arctan(y/a)+c\) but then you have to use integration by parts to integrate w.r.t x. so, i suggest we better CHANGE the ORDER of integration, that is, we'll change the limits so that we can first integrate w.r.t 'x' and then 'y'. For that, wee need to plot 2 regions (for 2 limits): (1) y= 0 to y=x^2/2 (2) x=0 to x= 2 can you plot these 4 curves (3 of these are simple straight lines)?
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i'm sorry don't know how to plot this..
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i'm sorry i forgot our previous lessons about that..
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
so, you say you don't know how to plot lines x=0,y=0,x=2 ? i'll help you with y=x^2/2 but if you don't know how to plot those lines, then it'll be very difficult for you to understand this..
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
so, plot these x=0,y=0,x=2 using 'Draw' tool here.
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
where is the line x=2 ?
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
sorry i forgot x = 2..
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
yes, x=0 is yaxis, y=0 is x axis , and x=2 you have drawn correct, now y= x^2/2 is a parabola which has a plot, dw:1362400675976:dw so you are integrating in the region that i am gonna shade
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
why it becomes parabola?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
dw:1362400803859:dw since we were integrating w.r.t 'y' first, i've drawn vertical lines, now we are changing the order of integration, we need to draw horizontal lines,
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
dw:1362400889186:dw to dw:1362400935932:dw
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i'm sorry i forgot how to find the equation of the line
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
oh, its easy, that a horizontal line , so u just need coordinates of intersection, which you can find by solving simultaneously, x=2 and x= \(\sqrt{2y}\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
by the way, did you get that final diagram ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
which part you have doubt with ?
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
\[x =2 and x = \sqrt{2y}\] i don't know how to fnd the equation of this line..
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
x= 2, x=\(\sqrt{2y}\) so, 2= \(\sqrt{2y}\) can u find 'y' from here ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
:O that was simple arithmetic....sorry to say but, if you have problem solving that you should think before dealing double integration, sorry if i am rude, but you are lacking basics ... 2= \(\sqrt{2y}\) 4= 2y (by squaring) y=2 this is the equation of your horizontal line
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
yes that's true i almost forgot the topics about integration that's why i am like this..
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
dw:1362402005207:dw so your new limits are, \(\huge \int \limits_0^2\int \limits_{\sqrt{2y}}^2\dfrac{x}{\sqrt{1+x^2+y^2}}dx.dy \) now you have to first integrate w.r.t x to do that put u = 1+x^2+y^2 du=... ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
oh, since we are integrating w.r.t 'x' first now, we treat 'y' as constant, so du =2xdx or x dx = du/2 got this ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
so, your inner integral will become, \(\int \dfrac{du}{2\sqrt u}\) can you integarte this ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
as long as weve gotten this u substitution correct: \[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
assuming that u = 1+x^2+y^2, plug back in for u \[\sqrt{ u}=\sqrt{ 1+x^2+y^2}\] from x=sqrt(2y) to 2 \[\sqrt{ 1+2^2+y^2}\sqrt{ 1+(\sqrt{2y})^2+y^2}\] \[\sqrt{ 5+y^2}\sqrt{ 1+2y+y^2}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
y^2+2y+1 = (y+1)^2 sooo \[\int\sqrt{5+y^2}(y+1)~dy\]
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
why it becomes subtraction?
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
why it become y^2+2y+1 = (y+1)^2 ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
because: \(\Large \int_{a}^{b} f(x)~dx=F(b)F(a)\)
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
in this time we are integrating y?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
when we replace the \(\sqrt{u}\) with what we defined it to begin with: \(u=1+x^2+y^2\) and input the ab limits of a=\(\sqrt{2y}\) and b=2 we get \[(\sqrt{1+b^2+y^2})(\sqrt{1+a^2+y^2})\] \[(\sqrt{1+b^2+y^2})(\sqrt{1+a^2+y^2})\] and yes, then since we are finished with the inside integral, we focus on the next integral
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the next integration is next ... \[\int\sqrt{5+y^2}(y+1)~dy\]
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
is it correct?i'm sorry have a problem in integration..
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the square root part seems a little off to me
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
and it looks like you tried a derivative onthe y+1 parts instead of integeration
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\int y+1~dy=(\frac{y^2}{2}+y)\]
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
yes.. i'm sorry a have a problem in integration..
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
why is that u not integrate \[\sqrt{5 + y^2}\]?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i did not integrate it because I want you to work it out. you are the one that needs the practice at it :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
if you cant see the integration right off hand, try the "table" method ... which is simply looking up the integral in a table of integrals: look for \(\sqrt{a^2+x^2}\)
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i looked it up in a table, just to be sure. Look at #30
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\int \sqrt{5+y^2}dy=\frac12y\sqrt{5+y^2}+\frac12 5^2ln(y+\sqrt{5+y^2})\]from y=0 to y=2
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the 5^2 ln part is a typo; just 5 ln ...
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
\[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
plug in the values of your limit to determine the values
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[(3+\frac52 ln(5))(\frac54ln(5))4\] the 4 is from the simpler part we did earlier
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
with any luck this simplifies to:\[\frac54ln(5)1\]
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
is this the answer? why 4 ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\int\sqrt{5+y^2}(y+1)~dy\] \[\int\sqrt{5+y^2}dy\int y+1~dy\] \[\int\sqrt{5+y^2}dy(\frac12y^2+y)\] and since y=0 to 2 \[\int\sqrt{5+y^2}dy4\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i cant verify the steps taken to swap the limits from dy dx to dx dy at the moment, but assuming that was done correctly .... we have come to the end of it
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i'm sorry..i don't understand it. it's a little bit difficult... :(
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the hardest part was not the y^2/2+y ... that just equals 4 we worked thru the sqrt(5+y^2) from the table, and inputted the limit values to get: \[\frac54ln(5)4\]as a final result, assuming the swapping of dydx to dxdy was correct to begin with
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
i must study first and review about integral in order for me to understand this..
 one year ago

jedai17Best ResponseYou've already chosen the best response.0
please teach me again next time..thanks for your time.. :)
 one year ago
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