anonymous
  • anonymous
what is double integral function? how to solve double integral function?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
is the value of dy similar to the value of dx?
anonymous
  • anonymous
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.

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anonymous
  • anonymous
ahh..ok tnx..can you teach me how to solve this?i have no background in solving this,, \[\int\limits_{0}^{2}\int\limits_{0}^{x ^{2}/2} x/\sqrt{1 + x^2 + y^2} dydx\]
hartnn
  • hartnn
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 ?
anonymous
  • anonymous
why it become a^2?
hartnn
  • hartnn
i just took 1+x^2 = constant = a^2 because we have a standard integration formula for, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy\)
hartnn
  • hartnn
then we need to re-substitute back after integration, a^2= 1+x^2 before integrating w.r.t x
hartnn
  • hartnn
*after integration w.r.t y,
anonymous
  • anonymous
in this we are integrating y and we assume as x as constant?
anonymous
  • anonymous
what do you mean by w.r.t y, ?
hartnn
  • hartnn
yes, true, so we'll get answer in 'x' after first integrating w.r.t y (with respect to y)
hartnn
  • hartnn
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)?
anonymous
  • anonymous
i'm sorry don't know how to plot this..
anonymous
  • anonymous
i'm sorry i forgot our previous lessons about that..
hartnn
  • hartnn
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..
anonymous
  • anonymous
i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry
hartnn
  • hartnn
so, plot these x=0,y=0,x=2 using 'Draw' tool here.
anonymous
  • anonymous
|dw:1362400369683:dw|
anonymous
  • anonymous
is it correct?
hartnn
  • hartnn
where is the line x=2 ?
anonymous
  • anonymous
|dw:1362400538076:dw|
anonymous
  • anonymous
sorry i forgot x = 2..
hartnn
  • hartnn
yes, x=0 is y-axis, 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
anonymous
  • anonymous
why it becomes parabola?
hartnn
  • hartnn
|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,
hartnn
  • hartnn
|dw:1362400889186:dw| to |dw:1362400935932:dw|
anonymous
  • anonymous
i'm sorry i forgot how to find the equation of the line
hartnn
  • hartnn
oh, its easy, that a horizontal line , so u just need co-ordinates of intersection, which you can find by solving simultaneously, x=2 and x= \(\sqrt{2y}\)
hartnn
  • hartnn
by the way, did you get that final diagram ?
anonymous
  • anonymous
not yet..
hartnn
  • hartnn
which part you have doubt with ?
anonymous
  • anonymous
\[x =2 and x = \sqrt{2y}\] i don't know how to fnd the equation of this line..
hartnn
  • hartnn
x= 2, x=\(\sqrt{2y}\) so, 2= \(\sqrt{2y}\) can u find 'y' from here ?
anonymous
  • anonymous
no..
anonymous
  • anonymous
\[y=\sqrt{2}/2 ?\]
hartnn
  • hartnn
: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
anonymous
  • anonymous
yes that's true i almost forgot the topics about integration that's why i am like this..
hartnn
  • hartnn
|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=... ?
anonymous
  • anonymous
du = 2x + 2y
hartnn
  • hartnn
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 ?
anonymous
  • anonymous
yes..
hartnn
  • hartnn
so, your inner integral will become, \(\int \dfrac{du}{2\sqrt u}\) can you integarte this ?
anonymous
  • anonymous
1/2..is it correct?
amistre64
  • amistre64
as long as weve gotten this u substitution correct: \[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]
amistre64
  • amistre64
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 \[\sqrt{ 1+2^2+y^2}-\sqrt{ 1+(\sqrt{2y})^2+y^2}\] \[\sqrt{ 5+y^2}-\sqrt{ 1+2y+y^2}\]
amistre64
  • amistre64
y^2+2y+1 = (y+1)^2 sooo \[\int\sqrt{5+y^2}-(y+1)~dy\]
anonymous
  • anonymous
why it becomes subtraction?
anonymous
  • anonymous
why it become y^2+2y+1 = (y+1)^2 ?
amistre64
  • amistre64
because: \(\Large \int_{a}^{b} f(x)~dx=F(b)-F(a)\)
anonymous
  • anonymous
ah ok tnx..
anonymous
  • anonymous
in this time we are integrating y?
amistre64
  • amistre64
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
anonymous
  • anonymous
what's next?
amistre64
  • amistre64
the next integration is next ... \[\int\sqrt{5+y^2}-(y+1)~dy\]
anonymous
  • anonymous
\[\sqrt{5 + y} - y\]
anonymous
  • anonymous
is it correct?i'm sorry have a problem in integration..
amistre64
  • amistre64
the square root part seems a little off to me
amistre64
  • amistre64
and it looks like you tried a derivative onthe y+1 parts instead of integeration
amistre64
  • amistre64
\[-\int y+1~dy=-(\frac{y^2}{2}+y)\]
anonymous
  • anonymous
yes.. i'm sorry a have a problem in integration..
anonymous
  • anonymous
why is that u not integrate \[\sqrt{5 + y^2}\]?
amistre64
  • amistre64
i did not integrate it because I want you to work it out. you are the one that needs the practice at it :)
amistre64
  • amistre64
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 \(\sqrt{a^2+x^2}\)
anonymous
  • anonymous
ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?
amistre64
  • amistre64
i looked it up in a table, just to be sure. Look at #30
1 Attachment
amistre64
  • amistre64
a^2 = 5, and x = y
amistre64
  • amistre64
\[\int \sqrt{5+y^2}dy=\frac12y\sqrt{5+y^2}+\frac12 5^2ln(y+\sqrt{5+y^2})\]from y=0 to y=2
amistre64
  • amistre64
the 5^2 ln part is a typo; just 5 ln ...
anonymous
  • anonymous
\[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]
amistre64
  • amistre64
yes
anonymous
  • anonymous
what's next?
amistre64
  • amistre64
plug in the values of your limit to determine the values
amistre64
  • amistre64
\[(3+\frac52 ln(5))-(\frac54ln(5))-4\] the -4 is from the simpler part we did earlier
amistre64
  • amistre64
with any luck this simplifies to:\[\frac54ln(5)-1\]
anonymous
  • anonymous
is this the answer? why -4 ?
amistre64
  • amistre64
\[\int\sqrt{5+y^2}-(y+1)~dy\] \[\int\sqrt{5+y^2}dy-\int y+1~dy\] \[\int\sqrt{5+y^2}dy-(\frac12y^2+y)\] and since y=0 to 2 \[\int\sqrt{5+y^2}dy-4\]
amistre64
  • amistre64
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
anonymous
  • anonymous
is that the answer?
amistre64
  • amistre64
is what the answer?
anonymous
  • anonymous
i'm sorry..i don't understand it. it's a little bit difficult... :(
amistre64
  • amistre64
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
anonymous
  • anonymous
i must study first and review about integral in order for me to understand this..
amistre64
  • amistre64
good luck :)
anonymous
  • anonymous
please teach me again next time..thanks for your time.. :)

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