what is the best way to solve a double integral question without drawing the figure we need to integrate over?

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what is the best way to solve a double integral question without drawing the figure we need to integrate over?

Mathematics
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guessing lol
havent tried doubles yet....
lol.. but you said you were guessing..

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Other answers:

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

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