Evaluating a double integral

Question

I have to evaluate this double integral:

$$\int_0^1\int_0^1\cos\ (\max \ \{x^3,y^{\frac{3}{2}} \} )\ dxdy$$

I have hint with me that this is to be done with help of Greens theorem but i dont know how to start it

Please help me with this.

Thanks

In any case, start dividing the integration domain along the curve $x^3=y^{3/2}$. — Martín-Blas Pérez Pinilla, Oct 10 '14 at 09:38
in any case what do you mean by this . thanks pls be more specific i am kind of weak student in math — EulerRamanujan121, Oct 10 '14 at 10:09
"In any case" is with or without Green.Your big problem is that $\max$ is a piecewise function. In each piece, $\max(\cdots)$ will be simply $x^3$ or $y^{3/2}$. — Martín-Blas Pérez Pinilla, Oct 10 '14 at 10:15

score 2 · Accepted Answer · answered Oct 13 '14 at 06:23

2

Hint: divide the integration domain along the curve $x^3=y^{3/2}$. In each subdomain the integrand is function of only one variable: $$\int_0^1\int_0^1\cos(\max \{x^3,y^{\frac{3}{2}} \} )\,dxdy= \iint_{D_1}\cos(x^3)\,dxdy+\iint_{D_2}\cos(y^{3/2})\,dxdy. $$ Can you continue?

answered Oct 13 '14 at 06:23

Martín-Blas Pérez Pinilla

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is d1 THE SIDE OF SQUARE FROM (O,O) TO (1,0) AND D2 FROM (1,0) TO (1,1) . IF YES then how come u have taken x^3 for former piece. thnx for help n sorry for being dumb – EulerRamanujan121 Oct 13 '14 at 08:51
$D1$, $D2$ are 2D chunks of the square. The border of $D1$ is the line segment from $(0,0)$ to $(1,0)$ $\cup$ the line segment from $(1,0)$ to $(1,1)$ $\cup$ the curve $x^3=y^{3/2}$ from $(1,1)$ to (0,0). When $(x,y)\in D_1$, $x^3\ge y^{3/2}$ and $\max(x^3,y^{3/2})=\cdots$ – Martín-Blas Pérez Pinilla Oct 13 '14 at 09:02
ok i have quite understodd you . u mean dividing square into two triangles u have written d1 for one of triangle and then evaluating region over triangles(both 1 by 1 ) . my final question is how can u say that for D1 x^3 > y^3/2 mathematically ! thanks a lot – EulerRamanujan121 Oct 13 '14 at 09:10
(a) triangles with a non-straight side. (b) re-write $x^3=y^{3/2}$ as $y=\cdots$ – Martín-Blas Pérez Pinilla Oct 13 '14 at 09:21
can we check this by making graphs of x^3 and x^3/2 in d1 and see which is up amd that vl be maximum ? or is there another way – EulerRamanujan121 Oct 13 '14 at 09:21
There is only one graph. – Martín-Blas Pérez Pinilla Oct 13 '14 at 09:21
graph is square only ..but im saying that there is max function involved and for region D1 what output vl max function gives how to find that is my question really
– EulerRamanujan121 Oct 13 '14 at 09:25
I mean the graph of $x^3=y^{3/2}$. About the inequality, a easier question: what happens above and below $y=x$? If you understand this, you will understand why $x^3\ge y^{3/2}$ in $D_1$. – Martín-Blas Pérez Pinilla Oct 13 '14 at 09:36
Thanks below y=x x>y and then radicals vl take care of rest stuff . thanks buddy – EulerRamanujan121 Oct 13 '14 at 09:52

Evaluating a double integral

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