Skip to main content
Contents Index
Calc
Dark Mode Prev Up Next
\(\newcommand{\R}{\mathbb{R}}
\newcommand{\va}{\vec{a}}
\newcommand{\vb}{\vec{b}}
\newcommand{\vc}{\vec{c}}
\newcommand{\vC}{\vec{C}}
\newcommand{\vd}{\vec{d}}
\newcommand{\ve}{\vec{e}}
\newcommand{\cursedihat}{\hat{\dot{i}}}
\newcommand{\vi}{\hat{\imath}}
\newcommand{\vj}{\hat{\jmath}}
\newcommand{\vk}{\hat{k}}
\newcommand{\vn}{\vec{n}}
\newcommand{\vm}{\vec{m}}
\newcommand{\vr}{\vec{r}}
\newcommand{\vs}{\vec{s}}
\newcommand{\vu}{\vec{u}}
\newcommand{\vv}{\vec{v}}
\newcommand{\vw}{\vec{w}}
\newcommand{\vx}{\vec{x}}
\newcommand{\vy}{\vec{y}}
\newcommand{\vz}{\vec{z}}
\newcommand{\vzero}{\vec{0}}
\newcommand{\vF}{\vec{F}}
\newcommand{\vG}{\vec{G}}
\newcommand{\vH}{\vec{H}}
\newcommand{\vR}{\vec{R}}
\newcommand{\vT}{\vec{T}}
\newcommand{\vN}{\vec{N}}
\newcommand{\vL}{\vec{L}}
\newcommand{\vB}{\vec{B}}
\newcommand{\vS}{\vec{S}}
\newcommand{\proj}{\text{proj}}
\newcommand{\comp}{\text{comp}}
\newcommand{\nin}{}
\newcommand{\vecmag}[1]{\left\lVert #1\right\rVert}
\newcommand{\grad}{\nabla}
\newcommand\restrict[1]{\raise-.5ex\hbox{$\Big|$}_{#1}}
\DeclareMathOperator{\curl}{curl}
\DeclareMathOperator{\divg}{div}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Activity 12.4.3 .
Suppose we want to find the area of the bounded region \(D\) between the curves
\begin{equation*}
y = 1-x^2 \quad \quad \text{ and } \quad \quad y=x-1
\end{equation*}
Figure 12.4.2. The graphs of \(y = 1-x^2\) and \(y=x-1\)
(a)
The volume of a solid with constant height is given by the area of the base times the height. Hence, we may interpret the area of the region
\(D\) as the volume of a solid with base
\(D\) and of uniform height 1. That is, the area of
\(D\) is given by
\(\iint_D 1 \, dA\text{.}\) Write an iterated integral whose value is
\(\iint_D 1 \, dA\text{.}\)
Hint .
Which order of integration might be more efficient? Why?
(b)
Evaluate the iterated integral from (a). What does the result tell you?