exercise:8fd9ac3ad0: Difference between revisions
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in terms of <math>x</math> that the graph of the function | 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. | <math>f(x) = \sqrt{a^2-(x-h)^2}</math> is a semicircle. | ||
<ul style{{=}}"list-style-type:lower-alpha" | <ul style{{=}}"list-style-type:lower-alpha"> | ||
<li> | <li>Assuming that <math>h > a</math> and using the method | ||
Assuming that <math>h > a</math> and using the method | |||
of cylindrical shells, write a definite integral | of cylindrical shells, write a definite integral | ||
for the volume of the solid torus (doughnut) | for the volume of the solid torus (doughnut) | ||
with radii <math>h</math> and <math>a</math>.</li> | with radii <math>h</math> and <math>a</math>.</li> | ||
<li>Evaluate the integral in | <li>Evaluate the integral in (a) by | ||
making the change of variable <math>y=x-h</math> | making the change of variable <math>y=x-h</math> | ||
, | , |
Latest revision as of 01:16, 25 November 2024
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[/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.
- 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 (a) 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).