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we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes fromtill
now we what first integrate by y then by x
so when we integrate by y: our intervals aretill
what are the intervals when we integrate by x
?
we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from
now we what first integrate by y then by x
so when we integrate by y: our intervals are
what are the intervals when we integrate by x
?
Did you draw the region defined by the terminals of the given double integral? It should then be obvious.Originally Posted by transgalactic
we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from
now we what first integrate by y then by x
so when we integrate by y: our intervals are
what are the intervals when we integrate by x
?
Your terminals tell you
we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from
now we what first integrate by y then by x
so when we integrate by y: our intervals are
what are the intervals when we integrate by x
?and
.
From the second inequality
and
.
So you have newterminals
.
Now from the first inequality
![\displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies y \leq \sqrt{x}](http://latex.codecogs.com/png.latex?\displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies y \leq \sqrt{x})
and
.
So you now haveterminals
.
Therefore
![\displaystyle \int_0^2{\int_{y^2}^{\sqrt[3]{y^2}}{y\,e^{x^2}\,dx}\,dy} = \int_0^{\sqrt[3]{4}}{\int_{\sqrt{x^3}}^{\sqrt{x}}{y\,e^{x^2}\,dy}\ ,dx}](http://latex.codecogs.com/png.latex?\displaystyle \int_0^2{\int_{y^2}^{\sqrt[3]{y^2}}{y\,e^{x^2}\,dx}\,dy} = \int_0^{\sqrt[3]{4}}{\int_{\sqrt{x^3}}^{\sqrt{x}}{y\,e^{x^2}\,dy}\ ,dx})
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