multivariable-calculus
Evaluate the double integrals by interation.
$\iint_{R}^{}xy^2dA$, where R is the finite region in the first quadrant bounded by the curves $y=x^2$ and $x=y^2$.
The answer is $3/56$.
My work:
The interception between $y=x^2$ and $x=y^2$ is (0,0) and (1,1).
$$\int_{0}^{1}(xy^2)dx=\left [ \frac{x^2y^2}{2} \right ]_{0}^{1}=\frac{y^2}{2}\\
\int_{0}^{1}(\frac{y^2}{2})dy=\left [ \frac{y^3}{6} \right ]_{0}^{1}=\frac{1}{6}$$
Well, obviously $\frac{1}{6}\neq\frac{3}{56}$. What have i done wrong?
If I well understand your question maybe that this intuitive interpretation can help.
When we calculate an area by a double integral, we subdivide this area in a sum of little ''area elements''. If we chose these elements as $dy dx$ or $dx dy$, this means that the elements are little rectangles of sides $dy$ and $dx$.
If we chose the order $dydx$ this means that we want to count these elements starting from ''vertical'' strips in which (in your case) the side $dy$ start from $2x^2$ and goes to $9-x^2$, so these values are the limits of the sum, and becomes the limits of the integration in $dy$, than we sum all these strips summing for the oter side $dx$ has limits $o$ and $\sqrt{3}$, and this gives the double integral: $$ \int_0^{\sqrt{3}}\int_{2x^2}^{9-x^2}dydx $$
If we change the order to $dx dy$ we count the area elements starting from horizontal stripes, and in this case the side $dx$ start from $x=0$ and goes to $\sqrt{y/2 }$ if $y\le 6$, and to $\sqrt{9-y }$ if $6<y\le 9$. So you have the other double integral that is divided in two parts.
I do not know what polar coordinates would give since you did not show your results and attempts.
As it is, we can compute the inner integral (this is the gaussian integral) and face the problem of $$\frac{\sqrt{\pi }}{2 \sqrt{a}}\int_{-d}^{+d}e^{\frac{ \left(b^2-4 a c\right)}{4 a}y^2} \left(\text{erf}\left(\frac{2 a d-b y}{2 \sqrt{a}}\right)+\text{erf}\left(\frac{2 a d+b y}{2 \sqrt{a}}\right)\right)\,dy$$ which does not seem to be possible.
Best Answer
It is true that the zone between the two curves is inside the square [0,1]x[0,1], but if you fix $x\in[0,1]$ then the $y$ 'moves' only between $x^{2}$ and $\sqrt{x}$ (you can check that easily by drawing the curves that delimit R), so the integral is
$$\int{\int_{R}{xy^{2}}}dA=\int_{0}^{1}({\int_{x^{2}}^{\sqrt{x}}{xy^{2}}dy})dx$$
The same can be done for $y\in[0,1]$, and then $x$ 'moves' between $y^{2}$ and $\sqrt{y}$, so the integral can be written
$$\int{\int_{R}{xy^{2}}}dA=\int_{0}^{1}({\int_{y^{2}}^{\sqrt{y}}{xy^{2}}dx})dy$$
(This can be done because of Fubini's theorem, and because the function is integrable in all of the intervals that appear above)
What you have done is the integral in all of the square [0,1]x[0,1], that's why the result is different.