Starting from the definition, calculate the following
Starting from the definition, calculate the following integral: $$ \iint_{\substack{a \leqslant x \leqslant b \\ c \leqslant y \leqslant d}} x^2 y^2 \, dx \, dy. $$
Explanation of Method
The integrand $f(x,y)=x^2y^2$ is separable: $f(x,y)=g(x)h(y)$ with $g(x)=x^2$, $h(y)=y^2$. We build the Riemann sum over $R=[a,b]\times[c,d]$, separate it into a product of two single sums, and pass to the limit so each sum becomes a single integral.
- Partition $[a,b]$ and $[c,d]$ into $n$ and $m$ pieces.
- Write the double Riemann sum for $x_i^2y_j^2$.
- Separate the sum into a product of an $x$-sum and a $y$-sum.
- Take the limit: each sum becomes $\int_a^b x^2\,dx$ and $\int_c^d y^2\,dy$.
- Evaluate and verify with iterated integrals and a special case.
Step 1 · Understanding the definition
$$ \iint_R f(x,y)\,dA = \lim_{n,m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m} f(x_i^*,y_j^*)\,\Delta x\,\Delta y, \qquad R=[a,b]\times[c,d],\ f(x,y)=x^2y^2. $$
Step 2 · Set up the partition
$$ \Delta x=\frac{b-a}{n}, \qquad \Delta y=\frac{d-c}{m}, $$
with right-endpoint sample points $x_i=a+i\frac{b-a}{n}$ ($i=1,\dots,n$) and $y_j=c+j\frac{d-c}{m}$ ($j=1,\dots,m$).
Step 3 · Write the Riemann sum
$$ S_{n,m}=\sum_{i=1}^{n}\sum_{j=1}^{m} x_i^2y_j^2\,\Delta x\,\Delta y = \frac{(b-a)(d-c)}{nm}\sum_{i=1}^{n}\sum_{j=1}^{m}\left(a+i\tfrac{b-a}{n}\right)^2\left(c+j\tfrac{d-c}{m}\right)^2. $$
Step 4 · Separate the sums
Since $x_i^2y_j^2=x_i^2\cdot y_j^2$:
$$ S_{n,m}=\frac{b-a}{n}\sum_{i=1}^{n} x_i^2 \cdot \frac{d-c}{m}\sum_{j=1}^{m} y_j^2. $$
Step 5 · Take the limit
$$ \lim_{n\to\infty}\frac{b-a}{n}\sum_{i=1}^{n} x_i^2 = \int_a^b x^2\,dx, \qquad \lim_{m\to\infty}\frac{d-c}{m}\sum_{j=1}^{m} y_j^2 = \int_c^d y^2\,dy. $$
Therefore $\displaystyle \iint_{\substack{a\leqslant x\leqslant b\\ c\leqslant y\leqslant d}} x^2y^2\,dx\,dy = \int_a^b x^2\,dx \cdot \int_c^d y^2\,dy$.
Step 6 · Evaluate the integrals
$$ \int_a^b x^2\,dx=\left[\frac{x^3}{3}\right]_a^b=\frac{b^3-a^3}{3}, \qquad \int_c^d y^2\,dy=\left[\frac{y^3}{3}\right]_c^d=\frac{d^3-c^3}{3}. $$
Step 7 · Multiply the results
$$ \iint_{\substack{a\leqslant x\leqslant b\\ c\leqslant y\leqslant d}} x^2y^2\,dx\,dy = \frac{b^3-a^3}{3}\cdot\frac{d^3-c^3}{3} = \frac{(b^3-a^3)(d^3-c^3)}{9}. $$
Step 8 · Verification using iterated integrals
$$ \int_a^b\int_c^d x^2y^2\,dy\,dx = \int_a^b x^2\cdot\frac{d^3-c^3}{3}\,dx = \frac{d^3-c^3}{3}\cdot\frac{b^3-a^3}{3} = \frac{(b^3-a^3)(d^3-c^3)}{9}. $$
This confirms our result. ✓
Step 9 · Alternative order of integration
$$ \int_c^d\int_a^b x^2y^2\,dx\,dy = \int_c^d y^2\cdot\frac{b^3-a^3}{3}\,dy = \frac{b^3-a^3}{3}\cdot\frac{d^3-c^3}{3} = \frac{(b^3-a^3)(d^3-c^3)}{9}. $$
Again we get the same result. ✓
Step 10 · Special case verification
For the unit square $a=0,b=1,c=0,d=1$:
$$ \iint_{[0,1]\times[0,1]} x^2y^2\,dx\,dy = \frac{(1-0)(1-0)}{9}=\frac19, $$
and directly, $\int_0^1 x^2\,dx\cdot\int_0^1 y^2\,dy=\frac13\cdot\frac13=\frac19$. Perfect match. ✓
Step 11 · Factorization using sum of cubes
Using $b^3-a^3=(b-a)(b^2+ab+a^2)$:
$$ \frac{(b^3-a^3)(d^3-c^3)}{9} = \frac{(b-a)(b^2+ab+a^2)(d-c)(d^2+cd+c^2)}{9}. $$
Summary
Key steps: recognize $f(x,y)=x^2y^2$ is separable; separate $S_{n,m}=\left[\frac{b-a}{n}\sum x_i^2\right]\cdot\left[\frac{d-c}{m}\sum y_j^2\right]$; take limits so each sum becomes an integral; evaluate $\frac{b^3-a^3}{3}\cdot\frac{d^3-c^3}{3}$; verify with iterated integrals in both orders.
Additional Notes
- For any separable function $f(x,y)=g(x)h(y)$, the double integral over a rectangle factors into the product of two single integrals — a very useful shortcut.
- The result represents the volume under $z=x^2y^2$ over the rectangle $[a,b]\times[c,d]$.
Result
The solution to the integral is: