exercise:04653989bc: Difference between revisions
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<math>a=0</math> and <math>b=\frac\pi2</math>.</li> | <math>a=0</math> and <math>b=\frac\pi2</math>.</li> | ||
<li><math>r=\sqrt{2\cos2\theta}</math>, <math>a=0</math> and <math>b=\frac\pi4</math>. | <li><math>r=\sqrt{2\cos2\theta}</math>, <math>a=0</math> and <math>b=\frac\pi4</math>. | ||
(See | (See [[guide:3cc850dbe4#exam 10.6.5|Example]])</li> | ||
</ul> | </ul> |
Latest revision as of 00:29, 26 November 2024
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[/math]
For each of the following equations [math]r=f(\theta)[/math] and pairs of numbers [math]a[/math] and [math]b[/math], draw the region [math]R[/math] consisting of all points with polar coordinates [math](r,\theta)[/math] such that [math]a\leq\theta\leq b[/math] and [math]0\leq r\leq f(\theta)[/math]. Compute [math]\mbox{''area''}(R)[/math].
- [math]r=4\sin\theta[/math], [math]a=0[/math] and [math]b=\pi[/math].
- [math]r=\frac4{\sin\theta}[/math], [math]a=\frac\pi4[/math] and [math]b=\frac{3\pi}4[/math].
- [math]r=2\theta[/math], [math]a=\pi[/math] and [math]b=2\pi[/math].
- [math]r=\frac1{2\cos\theta+3\sin\theta}[/math], [math]a=0[/math] and [math]b=\frac\pi2[/math].
- [math]r=\sqrt{2\cos2\theta}[/math], [math]a=0[/math] and [math]b=\frac\pi4[/math]. (See Example)