How to have this definite integral? 
Suppose that a function $f$ of $x$ and $y$ be defined as follows:$$f(x,y) =
\begin{cases}
\frac{21}{4}x^2y  & \text{for $x^2 \leq y\leq 1$,} \\
0 & \text{otherwise. }  \\
\end{cases}$$
  I have to determine the value of integral for which $y\leq x$ also holds.

The answer is $\frac{3}{20}$ and I also get it using figure for $x$ and $y$, but don't know how to get it with calculations.
 A: We find 
$$\int_{x=0}^1 \left(\int_{y=x^2}^{x}\frac{21}{4}x^2y\,dy\right)\,dx.$$
The integration with respect to $y$ gives $\frac{21}{8}x^2(x^2-x^4)$.
Integration with respect to $x$ yields $\frac{21}{8} \left(\frac{1}{5}-\frac{1}{7}\right)$. Simplify.
A: The given condition $y\le x$and $x^2\le y\le1$ implies $x^2\le y\le x$. 
Since $x^2\le x$, $0\le x\le 1$.
$$\int_{0}^1\int_{x^2}^x\frac{21}{4}x^2ydydx=\int_{0}^1\left[\frac{21}{8}x^2y^2\right]^x_{x^2}dx=\int_{0}^1\frac{21}{8}x^2(x^2-x^4)=\frac{21}{8}\left[\frac{x^5}{5}-\frac{x^7}{7}\right]^1_{0}=\frac{3}{20}$$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\begin{align}
{\cal I}
&
\equiv
\int_{-\infty}^{\infty}{\rm d}x\int_{-\infty}^{\infty}{\rm d}y\,
{21 \over 4}\,x^{2}y\,
\Theta\pars{y - x^{2}}\Theta\pars{1 - y}\Theta\pars{x - y}
\\[3mm]&=
{21 \over 4}\int_{-\infty}^{1}{\rm d}y\,y\quad
\overbrace{\int_{-\infty}^{\infty}{\rm d}x\,
x^{2}\,\Theta\pars{y - x^{2}}\Theta\pars{x - y}}^{\equiv\ {\cal J}}
\end{align}


\begin{align}
{\cal J}
&=
-\,{1 \over 3}\int_{-\infty}^{\infty}{\rm d}x\,
x^{3}\,\bracks{%
-2x\,\delta\pars{y - x^{2}}\Theta\pars{x - y}
+
\Theta\pars{y - x^{2}}\delta\pars{x - y}}
\\[3mm]&=
{2 \over 3}\int_{-\infty}^{\infty}{\rm d}x\,x^{4}\,\Theta\pars{y}\,
{\delta\pars{x - y^{1/2}} \over 2\verts{x}}\,\Theta\pars{y^{1/2} - y}
-
{1 \over 3}\,y^{3}\,\Theta\pars{y - y^{2}}
\\[3mm]&=
{1 \over 3}y^{3/2}\,\Theta\pars{y}\,\Theta\pars{y\bracks{1 -y}}
-
{1 \over 3}\,y^{3}\,\Theta\pars{y - y^{2}}
=
{1 \over 3}\pars{y^{3/2} - y^{3}}\Theta\pars{y}\Theta\pars{1 - y}
\end{align}


\begin{align}
{\cal I}
&=
{7 \over 4}\int_{0}^{1}\pars{y^{5/2} - y^{4}}\,{\rm d}y
=
{7 \over 4}\pars{{2 \over 7} - {1 \over 5}}
=
\color{#ff0000}{\large{3 \over 20}}
\end{align}

