what is double integral function?
how to solve double integral function?

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- anonymous

what is double integral function?
how to solve double integral function?

- chestercat

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- 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

is the value of dy similar to the value of dx?

- 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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## More answers

- 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

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

why it become a^2?

- 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

then we need to re-substitute back after integration, a^2= 1+x^2 before integrating w.r.t x

- hartnn

*after integration w.r.t y,

- anonymous

in this we are integrating y and we assume as x as constant?

- anonymous

what do you mean by w.r.t y, ?

- hartnn

yes, true, so we'll get answer in 'x' after first integrating w.r.t y (with respect to y)

- 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

i'm sorry don't know how to plot this..

- anonymous

i'm sorry i forgot our previous lessons about that..

- 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

i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry

- hartnn

so, plot these x=0,y=0,x=2 using 'Draw' tool here.

- anonymous

|dw:1362400369683:dw|

- anonymous

is it correct?

- hartnn

where is the line x=2 ?

- anonymous

|dw:1362400538076:dw|

- anonymous

sorry i forgot x = 2..

- 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

why it becomes parabola?

- 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

|dw:1362400889186:dw|
to
|dw:1362400935932:dw|

- anonymous

i'm sorry i forgot how to find the equation of the line

- 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

by the way, did you get that final diagram ?

- anonymous

not yet..

- hartnn

which part you have doubt with ?

- anonymous

\[x =2 and x = \sqrt{2y}\] i don't know how to fnd the equation of this line..

- hartnn

x= 2, x=\(\sqrt{2y}\)
so, 2= \(\sqrt{2y}\)
can u find 'y' from here ?

- anonymous

no..

- anonymous

\[y=\sqrt{2}/2 ?\]

- 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

yes that's true i almost forgot the topics about integration that's why i am like this..

- 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

du = 2x + 2y

- 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

yes..

- hartnn

so, your inner integral will become,
\(\int \dfrac{du}{2\sqrt u}\)
can you integarte this ?

- anonymous

1/2..is it correct?

- amistre64

as long as weve gotten this u substitution correct:
\[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]

- 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

y^2+2y+1 = (y+1)^2 sooo
\[\int\sqrt{5+y^2}-(y+1)~dy\]

- anonymous

why it becomes subtraction?

- anonymous

why it become y^2+2y+1 = (y+1)^2 ?

- amistre64

because: \(\Large \int_{a}^{b} f(x)~dx=F(b)-F(a)\)

- anonymous

ah ok tnx..

- anonymous

in this time we are integrating y?

- 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

what's next?

- amistre64

the next integration is next ...
\[\int\sqrt{5+y^2}-(y+1)~dy\]

- anonymous

\[\sqrt{5 + y} - y\]

- anonymous

is it correct?i'm sorry have a problem in integration..

- amistre64

the square root part seems a little off to me

- amistre64

and it looks like you tried a derivative onthe y+1 parts instead of integeration

- amistre64

\[-\int y+1~dy=-(\frac{y^2}{2}+y)\]

- anonymous

yes.. i'm sorry a have a problem in integration..

- anonymous

why is that u not integrate \[\sqrt{5 + y^2}\]?

- 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

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

ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?

- amistre64

i looked it up in a table, just to be sure. Look at #30

##### 1 Attachment

- amistre64

a^2 = 5, and x = y

- 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

the 5^2 ln part is a typo; just 5 ln ...

- anonymous

\[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]

- amistre64

yes

- anonymous

what's next?

- amistre64

plug in the values of your limit to determine the values

- amistre64

\[(3+\frac52 ln(5))-(\frac54ln(5))-4\]
the -4 is from the simpler part we did earlier

- amistre64

with any luck this simplifies to:\[\frac54ln(5)-1\]

- anonymous

is this the answer? why -4 ?

- 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

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

is that the answer?

- amistre64

is what the answer?

- anonymous

i'm sorry..i don't understand it. it's a little bit difficult... :(

- 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

i must study first and review about integral in order for me to understand this..

- amistre64

good luck :)

- anonymous

please teach me again next time..thanks for your time.. :)

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