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ViPhy
 one year ago
Best ResponseYou've already chosen the best response.0Double 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.

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0is the value of dy similar to the value of dx?

yunus
 one year ago
Best ResponseYou've already chosen the best response.0yes 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.

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0ahh..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
 one year ago
Best ResponseYou've already chosen the best response.0if 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 ?

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

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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0in this we are integrating y and we assume as x as constant?

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean by w.r.t y, ?

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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0ok, \(\int \dfrac{1}{\sqrt{a^2+y^2}}dy= (1/a)\arctan(y/a)+c\) 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)?

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry don't know how to plot this..

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry i forgot our previous lessons about that..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0so, 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..

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry

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

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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0sorry i forgot x = 2..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0why it becomes parabola?

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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry i forgot how to find the equation of the line

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

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

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

jedai17
 one year ago
Best 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..

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

hartnn
 one year ago
Best 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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0yes that's true i almost forgot the topics about integration that's why i am like this..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0dw: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=... ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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 ?

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1as long as weve gotten this u substitution correct: \[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1assuming 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
 one year ago
Best ResponseYou've already chosen the best response.1y^2+2y+1 = (y+1)^2 sooo \[\int\sqrt{5+y^2}(y+1)~dy\]

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0why it becomes subtraction?

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0why it become y^2+2y+1 = (y+1)^2 ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1because: \(\Large \int_{a}^{b} f(x)~dx=F(b)F(a)\)

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0in this time we are integrating y?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the next integration is next ... \[\int\sqrt{5+y^2}(y+1)~dy\]

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0is it correct?i'm sorry have a problem in integration..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the square root part seems a little off to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1and it looks like you tried a derivative onthe y+1 parts instead of integeration

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int y+1~dy=(\frac{y^2}{2}+y)\]

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0yes.. i'm sorry a have a problem in integration..

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0why is that u not integrate \[\sqrt{5 + y^2}\]?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i did not integrate it because I want you to work it out. you are the one that needs the practice at it :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if 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 \(\sqrt{a^2+x^2}\)

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i looked it up in a table, just to be sure. Look at #30

amistre64
 one year ago
Best 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 5^2 ln part is a typo; just 5 ln ...

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0\[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1plug in the values of your limit to determine the values

amistre64
 one year ago
Best 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1with any luck this simplifies to:\[\frac54ln(5)1\]

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0is this the answer? why 4 ?

amistre64
 one year ago
Best 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 \[\int\sqrt{5+y^2}dy4\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry..i don't understand it. it's a little bit difficult... :(

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 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

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0i must study first and review about integral in order for me to understand this..

jedai17
 one year ago
Best ResponseYou've already chosen the best response.0please teach me again next time..thanks for your time.. :)
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