Question

hello everyone .. what can you say about the double integral?

Answer 1

Fun.

Answer 2

can you explain what is double integral?

Answer 3

So if a single integral gives you the area under a curve, then a double integral would give you the volume under a surface, similar to how L*W=A and L*W*H=V

Answer 4

Except now it's, f(x)*dx=A and f(x,y)*dx*dy=V

Answer 5

Well, that would tell you the area and volume of an infinitesimally small piece, integration is the shortcut to adding up the infinite number of infinitely small areas and volumes from one point to the other, kind of like how a derivative is the shortcut to taking the limit of the slope , which amounts to practically dividing by zero... yup.

Answer 6

How do you find the double integral?

Answer 7

Do you have an example? You integrate with respect to one variable and then the other while holding the other one constant.

Answer 8

yes i have..can you answer this question \[\int\limits_{0}^{2}\int\limits_{0}^{x/2}\frac{ x }{ \sqrt{1+x^2+y^2} }\]

Answer 9

\[Evaluate : \int\limits^2_{y=1} \int\limits^3_{x=0}(1+8xy)dxdy Solution : In this example the inner \int\limits is \int\limits^3_{x=0}(1+8xy)dxdy with y treated as constant. \therefore : \int\limits^2_{y=1}[ \int\limits^3_{x=0}(1+8xy)dx]dy =\int\limits^2_{y=1} [ x + \frac{8x^2y}{2}]^3_{x=0}dy = \int\limits^2_{y=1} (3+36y ) dy = [3y+\frac{36y^2}{2}]^2_{y=1} = (6+72)-(3+18)= 57\]

Answer 10

Evaluate : \[\int\limits^2_{y=1} \int\limits^3_{x=0}(1+8xy)dxdy\]
In this example the inner integral is \[\int\limits^3_{x=0}(1+8xy)dxdy\] with y treated as constant.
\[\int\limits^2_{y=1}[ \int\limits^3_{x=0}(1+8xy)dx]dy =\int\limits^2_{y=1} [ x + \frac{8x^2y}{2}]^3_{x=0}dy\]
= \[\int\limits^2_{y=1} (3+36y ) dy \]
\[=[3y+\frac{36y^2}{2}]^2_{y=1} = (6+72)-(3+18)= 57\]

Answer 11

ow you give your own example.. what can you say about my example can you answer that ?

Answer 12

about that first we integrate dy \[\int\limits_{0}^{x/2} \frac{ x }{ \sqrt{1+x^2+y^2} }\]
= \[\int\limits_{0}^{x/2} x (1+x^2+y^2)\]
=x\[\int\limits_{0}^{x/2} \frac{ (1+x^2+y^2)^{-1/2} }{ 1/2 }\]
then integrating you will get this

\[(x ^{3}+x^5+\frac{ x^7 }{ 12 })^{1/2}\]
since you get the first integral
now the second integral

\[\int\limits_{0}^{2} x^3+x^5+\frac{ x^7 }{ 12 }\]

=\[\int\limits_{0}^{2} \frac{ (x^3+x^5+\frac{ x^7 }{ 12 })^3/2 }{ 3/2 }\]

after we integrating we will substituting the value of x



=(1664/144)^3/2
=(11.55555556)^3/2
=\[\sqrt{(11.55555556)^3}\]=39.28