Iterated Integrals over Rectangles

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Examples of Iterated Integrals

Example 1: (from the video) Compute $\displaystyle\iint_R(x-y)^2\,dA$ over the region with $x$ in $[0,2]$ and $y$ in $[0,1]$. As we will soon see, by Fubini's Theorem, $dA=dx\,dy$ or $dA=dy\,dx$; i.e. we can fix $x$ and integrate with respect to $y$ first, or vice versa.

Solution 1: We will compute $\displaystyle\iint_R(x-y)^2\,dA=\int_0^2\int_0^1 (x-y)^2\,dy\,dx=\int_0^2\left(\int_0^1 (x-y)^2\,dy\right)\,dx$.

DO: Integrate the parenthetical integral $\displaystyle\int_0^1 (x-y)^2\,dy$, treating $x$ as a constant, before reading more.

First we compute the antiderivative, then evaluate it (if you can do this substitution in your head, you can just evaluate the integral). Remember, $y$ is our variable and $x$ is a constant.
$\displaystyle\int (x-y)^2\,dy\overset{\fbox{$ \,\,u\,=\,x-y\\ du\,=\,(-1)\,dy$}\\}{=} -\int u^2\,du=-\frac{1}{3}u^3+c $, so $\displaystyle\int_0^1 (x-y)^2\,dy=-\frac{1}{3}(x-y)^3 \left|\begin{array}{c} ^1 \\ _0\end{array}\right .=-\frac{1}{3}\left((x-1)^3-(x-0)^3\right)=\frac{x^3-(x-1)^3}{3}$

DO: Integrate the outside integral, with the integrand being our answer from above: $\displaystyle\int_0^2\frac{x^3-(x-1)^3}{3}\,dx$ before reading more.

Here, we can see our antiderivative (after a mental substitution $u=x-1$, where $du=dx$)

$\displaystyle\int_0^2\frac{x^3-(x-1)^3}{3}\,dx=\frac{1}{3}\left(\frac{x^4}{4}-\frac{(x-1)^4}{4}\right)\left|\begin{array}{c} ^2 \\ _0 \end{array}\right .=\frac{1}{3}\left(\left(4-\frac{1}{4}\right)-\left(0-\frac{1}{4}\right)\right)=\frac{4}{3}$

In these two steps, one variable at a time, we have found $\displaystyle\iint_R(x-y)^2\,dA=\int_0^2\left(\int_0^1 (x-y)^2\,dy\right)\,dx=\int_0^2\left(\frac{x^3-(x-1)^3}{3}\right)\,dx=\frac{4}{3}$

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Example 2: Find the volume of the solid $W$ under the hyperbolic paraboloid $$z=f(x, y) = 2+ x^2 - y^2 $$ and over the square $D\,= \,[-1,\,1]\,\times\,[-1,\,1]$.

DO: Compute this integral before reading more. This is the graph of $f$:

Solution 2: We will compute $\displaystyle\int_{-1}^1\int_{-1}^1\, (2+x^2 - y^2)\, dy\,dx$.

$\displaystyle\int_{-1}^1\left(\int_{-1}^1\, (2+x^2 - y^2)\, dy\right)\,dx=\int_{-1}^1\left[\,2y +x^2y -\frac{y^3}{3}\right]_{-1}^1\,dx$ (after integration with respect to $y$, leaving $x$ constant)

$\displaystyle=\int_{-1}^1\left[(2+x^2-1/3)-(-2-x^2+1/3)\right]\,dx =\int_{-1}^1\left(\frac{10}{3} +2x^2\right)\,dx$ (and now integrate with respect to $x$)

$\displaystyle=\left(\frac{10}{3}x+\frac{2}{3}x^3\right)\left|\begin{array}{c} ^1 \\ _{-1} \end{array}\right .=\frac{10}{3}+\frac{2}{3}-\left(-\frac{10}{3}-\frac{2}{3}\right)=8$

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Examples of Iterated Integrals