anonymous
  • anonymous
what is the best way to solve a double integral question without drawing the figure we need to integrate over?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
guessing lol
amistre64
  • amistre64
havent tried doubles yet....
anonymous
  • anonymous
lol.. but you said you were guessing..

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amistre64
  • amistre64
whats the function or functions to double integrate :)
anonymous
  • anonymous
yeah giv us the function..
amistre64
  • amistre64
and make it a hard one lol
amistre64
  • amistre64
something with an exponent of 3 in it ;)
amistre64
  • amistre64
and a negative
anonymous
  • anonymous
let it be a double integral of 4xy-y^3 over the area bounded by the curves y=x^2 n y=x^(1/2)
amistre64
  • amistre64
if we do the x we get......umm........how do we do doubles anyways?
amistre64
  • amistre64
is this confined to a single plane right?
amistre64
  • amistre64
i see the football shape made by the bounds of x^2 and sqrt(x)
anonymous
  • anonymous
the region bounded by those curves looks like a rugby ball..
amistre64
  • amistre64
dunno what to do with the 4xy-y^3 function tho..... how does that come into play?
amistre64
  • amistre64
if we int with respect to y it goes; 2xy^2-y^4/4 right?
amistre64
  • amistre64
and with respect to x its: 4yx^3/3 +xy^3 ??
anonymous
  • anonymous
the major problem n which also the 1st step is to find the limits for the double integral for individual x and y.. once u r able to get that.. then its more or less like normal integration..
anonymous
  • anonymous
how hard is it to find limits solve sqrt(x) = x^2
anonymous
  • anonymous
x^4 = x x(x^3 -1) =0 x=0, x=1 for real solutions
anonymous
  • anonymous
when x=0 , y= 0 when x=1, y= 1
anonymous
  • anonymous
so its just \[\int\limits_{0}^{1}\int\limits_{0}^{1} f(x,y) dx dy \]
anonymous
  • anonymous
doesnt matter which order u integrate, in the above I do x first
anonymous
  • anonymous
so f(x,y) = 4xy-y^3 int f dx = 2x^2 y -xy^3 [ from x=1..x=0] = ( [2y -y^3 ] - [0] ) = 2y -y^3 now we integrate that with respect to y , from y=0 to y=1
anonymous
  • anonymous
so we get [ y^2 -y^4 / 4 ] from y=1..y=0 = 1- (1/4) = 3/4 //final answer

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