[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

Since [math](x-h)^2 + y^2 = a^2[/math] is an equation of the circle with radius [math]a[/math] and center at [math](h,0)[/math], it follows by solving for [math]y[/math] in terms of [math]x[/math] that the graph of the function [math]f(x) = \sqrt{a^2-(x-h)^2}[/math] is a semicircle.

• lab{8.4.11a} Assuming that [math]h \gt a[/math] and using the method of cylindrical shells, write a definite integral for the volume of the solid torus (doughnut) with radii [math]h[/math] and [math]a[/math].
• Evaluate the integral in \ref{ex8.4.11a} by making the change of variable [math]y=x-h[/math] , and using the fact that [math]\int_{-a}^a \sqrt{a^2-y^2} \; dy = \frac{\pi a^2}2[/math] (area of a semicircle).