683
Concepts and Vocabulary
The area \(A\) of the sector of a circle of radius \(r\) and central angle \(\theta \) is \(A=\) ______________.
True or False The area enclosed by the graph of a polar equation and two rays that have the pole as a common vertex is found by approximating the area using sectors of a circle.
True or False The area \(A\) enclosed by the graph of the equation \(r=f(\theta ),\) \(r\geq 0,\) and the rays \(\theta =\alpha \) and \(\theta =\beta ,\) is given by \(A=\int_{\alpha }^{\beta }f( \theta)\, d\theta .\)
True or False If \(x( \theta) =r\cos \theta ,\) \(y( \theta) =r\sin \theta \) are parametric equations of the polar equation \(r=f( \theta) ,\) then \(\left( \dfrac{dx}{ d\theta }\right) ^{2}+\left( \dfrac{dy}{d\theta }\right) ^{2}=r^{2}+\left( \dfrac{dr}{d\theta }\right) ^{2}.\)
Skill Building
In Problems 5–8, find the area of the shaded region.
\(r=\cos (2\theta)\)
\(r=2 \sin (3\theta)\)
\(r=2+2\sin \theta\)
\(r=3-3\cos \theta\)
In Problems 9–12, find the area of the region enclosed by the graph of each polar equation swept out by the given rays.
\(r=3\cos \theta;\;\theta =0\;\hbox{to}\;\theta =\dfrac{\pi}{3}\)
\(r=3\sin \theta ;\;\theta =0\;\hbox{to}\;\theta = \dfrac{\pi }{4}\)
\(r=a\,\theta ;\;\theta =0\ \hbox{ to }\ \theta =2\pi \)
\(r=e^{a\,\theta };\;\theta =0\ \hbox{ to }\ \theta =\dfrac{\pi }{2}\)
In Problems 13–18, find the area of the region enclosed by the graph of each polar equation.
\(r=1+\cos \theta \)
\(r=2-2\sin \theta \)
\(r=3+\sin \theta \)
\(r=3(2-\sin \theta )\)
\(r=8\sin ( 3\theta ) \)
\(r=\cos (4\theta ) \)
In Problems 19–22, find the area of the region enclosed by one loop of the graph of each polar equation.
\(r=4\sin ( 2\theta ) \)
\(r=5\cos ( 3\theta ) \)
\(r^{2}=4\cos ( 2\theta ) \)
\(r=a^{2}\cos ( 2\theta ) \)
In Problems 23–26, find the area of each region described.
Inside \(r=2\sin \theta \); outside \(r=1\)
Inside \(r=4\cos \theta \); outside \(r=2\)
Inside \(r=\sin \theta \); outside \(r=1-\cos \theta \)
Inside \(r^{2}=4\cos ( 2\theta ) \); outside \(r=\sqrt{2}\)
In Problems 27–30, find the surface area of the solid of revolution generated by revolving each curve about the polar axis.
\(r=\sin \theta ,\;0\leq \theta \leq \dfrac{\pi }{2}\)
\(r=1+\cos \theta ,\;0\leq \theta \leq \pi \)
\(r=e^{\theta },\;0\leq \theta \leq \pi \)
\(r=2a\cos \theta ,\;0\leq \theta \leq \dfrac{\pi }{2}\)
Applications and Extensions
In Problems 31–48, find the area of the region:
enclosed by the small loop of the limaçon \(r=1+2\cos \theta \).
enclosed by the small loop of the limaçon \(r=1+2\sin \theta \).
684
enclosed by the loop of the graph of \(r=2-\sec \theta \).
enclosed by the loop of the graph of \(r=5+\sec \theta \).
enclosed by \(r=2\sin ^{2}\dfrac{\theta }{2}\).
enclosed by \(r=6\cos ^{2}\theta \).
inside the circle \(r=8\cos \theta \) and to the right of the line \(r=2\sec \theta \).
inside the circle \(r=10\sin \theta \) and above the line \( r=2\csc \theta \).
outside the circle \(r=3\) and inside the cardioid \(r=2+2\cos \theta \).
inside the circle \(r=\sin \theta \) and outside the cardioid \( r=1+\cos \theta \).
common to the circle \(r=\cos \theta \) and the cardioid \( r=1-\cos \theta \).
common to the circles \(r=\cos \theta \) and \(r=\sin \theta \).
common to the inside of the cardioid \(r=1+\sin \theta \) and the outside of the cardioid \(r=1+\cos \theta \).
common to the inside of the lemniscate \(r^{2}=8\cos ( 2\theta ) \) and the outside of the circle \(r=2.\)
enclosed by the rays \(\theta =0\) and \(\theta =1\) and \( r=e^{-\theta },\;\ 0\leq \theta \leq 1\).
enclosed by the rays \(\theta =0\) and \(\theta =1\) and \( r=e^{\theta },\;\ 0\leq \theta \leq 1\).
enclosed by the rays \(\theta =1\) and \(\theta =\pi \) and \(r= \dfrac{1}{\theta },\;\ 1\leq \theta \leq \pi \).
inside the outer loop but outside the inner loop of \(r=1+2\sin \theta \).
Area Find the area of the loop of the graph of \(r=\sec \theta +2\).
Surface Area of a Sphere Develop a formula for the surface area of a sphere of radius \(R\).
Surface Area of a Bead A sphere of radius \(R\) has a hole of radius \(a\lt R\) drilled through it. See the figure. The axis of the hole coincides with a diameter of the sphere.
Surface Area of a Plug A plug is made to repair the hole in the sphere in Problem 51.
Area Find the area enclosed by the loop of the strophoid \(r=\sec \theta -2\cos \theta ,\;-\dfrac{\pi}{2}\lt\theta \lt \dfrac{\pi}{2}\) as shown in the figure.
Area
Challenge Problems
Show that the area enclosed by the graph of \(r\theta =a\) and the rays \(\theta =\theta _{1}\) and \(\theta =\theta _{2}\) is proportional to the difference of the radii, \(r_{1}-r_{2}\), where \(r_{1}=\dfrac{a}{\theta _{1}}\) and \(r_{2}=\dfrac{a}{\theta _{2}}.\)
Find the area of the region that lies outside the circle \(r=1\) and inside the rose \(r=3\sin ( 3\theta )\).
Find the area of the region that lies inside the circle \(r=2\) and outside the rose \(r=3\sin ( 2\theta ) \).