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Concepts and Vocabulary
Convert the function f(x,y)=√4x2+4y2 to polar coordinates.
f(r,θ)=2r
In polar coordinates, the differential dA of area is dA = _________.
rdrdθ
True or False Geometrically, if f(x,y)≥0 and if f is continuous over a closed, bounded region R, then the double integral ∬ represents the volume of the solid under the surface z=f(x,y) and above the region R in the xy-plane.
False
True or False \iint\limits_{\kern-3ptR}r\,dr\,d\theta equals the area A of the closed, bounded region R.
True
Skill Building
In Problems 5–8, assume that z=f(x,y) is a continuous function over the region R. Express the double integral \displaystyle\iint\limits_{\displaystyle\kern-3ptR}f(x,y) \, {\it dA} over R as an iterated integral using polar coordinates.
\int_0^{2\pi} \int_0^2 f(r \cos \theta, r \sin \theta) r \, dr\, d\theta
\int_0^{2\pi} \int_1^3 f(r \cos \theta, r \sin \theta) r \, dr\, d\theta
In Problems 9–14, express each double integral over the region R as an iterated integral using polar coordinates; do not integrate.
\iint\limits_{\kern-3ptR} ( x+y) \, {\it dA}, R is enclosed by x^{2}+y^{2}=9, x\geq 0, and y\geq 0.
\int_0^{\pi/2} \int_0^3 r^2 (\cos \theta + \sin \theta) \, dr \, d\theta
\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \, {\it dA}, R is enclosed by y=\sqrt{1-x^{2}} and the lines y=x and y=-x.
\iint\limits_{\kern-3ptR}x\, {\it dA}, R is enclosed by x=\sqrt{4-y^{2}}, the positive x-axis, and the line y=x.
\int_0^{\pi/4} \int_0^2 r^2 \cos \theta \, dr \, d\theta
\iint\limits_{\kern-3ptR}y\, {\it dA}, R is enclosed by y=\sqrt{16-x^{2}}, in the first quadrant.
\iint\limits_{\kern-3ptR}( x+y^{2}) \, {\it dA}, R is enclosed by x^{2}+y^{2}=1 and x^{2}+y^{2}=4.
\int_0^{2\pi} \int_1^2 (r^2 \cos \theta + r^3 \sin^2\theta) \, dr \, d\theta
\iint\limits_{\kern-3ptR}\sqrt{x^{2}+y^{2}}\, {\it dA}, R is enclosed by x^{2}+y^{2}=\dfrac{1}{4} and x^{2}+y^{2}=5.
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In Problems 15–22, find each iterated integral.
\displaystyle\int_{0}^{\pi /4}\int_{0}^{4\sin \theta }r\cos \theta \,{\it dr}\,d\theta
\dfrac{2\sqrt{2}}3
\displaystyle\int_{0}^{5\pi /3}\int_{0}^{2\cos \theta }r\sin \theta \,{\it dr}\,d\theta
\displaystyle\int_{0}^{\pi /2}\int_{0}^{\pi /3}r\,\,d\theta\, {\it dr}
\dfrac{\pi^3}{24}
\displaystyle\int_{0}^{\pi /3}\int_{0}^{\sin \theta }r\,{\it dr}\,d\theta
\displaystyle\int_{0}^{\pi /4}\int_{0}^{\cos \theta }r\,{\it dr}\,d\theta
\dfrac{\pi+2}{16}
\displaystyle\int_{0}^{4}\int_{-\pi /4}^{\pi /4}r\cos \theta \,d\theta\, {\it dr}
\displaystyle\int_{0}^{2}\int_{-\pi /3}^{5\pi /3}r\sin \theta \,d\theta\, {\it dr}
0
\displaystyle\int_{0}^{\pi}\int_{0}^{\pi /4}r\,\,d\theta\, {\it dr}
In Problems 23–28, graph the region R, then find each double integral.
\displaystyle\iint\limits_{\kern-3ptR}3\sin \theta \,{\it dA} ,\quad0\leq r\leq 2, 0\leq \theta \leq \dfrac{\pi }{2}
6
\displaystyle\iint\limits_{\kern-3ptR}4\cos \theta \,{\it dA} ,\quad 0\leq r\leq 3, 0\leq \theta \leq \dfrac{\pi }{2}
\displaystyle\iint\limits_{\kern-3ptR}2r \sin \theta \,{\it dA} ,\quad 0\leq r\leq 1, 0\leq \theta \leq \dfrac{\pi }{2}
\dfrac23
\displaystyle\iint\limits_{\kern-3ptR}3r \cos \theta \,{\it dA} ,\quad 0\leq r\leq 2, 0\leq \theta \leq \dfrac{\pi }{4}
\displaystyle\iint\limits_{\kern-3ptR}3r^{2}\sin \theta \,{\it dA} ,\quad 0\leq r\leq 2, -\dfrac{\pi }{2}\leq \theta \leq \dfrac{\pi }{2}
0
\displaystyle\iint\limits_{\kern-3ptR}2r^{2}\cos \theta \,{\it dA} ,\quad 0\leq r\leq 4, - \dfrac{\pi }{2}\leq \theta \leq \dfrac{\pi }{2}
In Problems 29–42, find each double integral by changing to polar coordinates.
\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-x^{2}}}\,{\it dy}\,{\it dx}
\dfrac{\pi}4
\displaystyle\int_{0}^{3}\int_{0}^{y}\sqrt{x^{2}+y^{2}}\,{\it dx}\,{\it dy}
\displaystyle\int_{-2}^{2}\int_{0}^{\sqrt{4-y^{2}}}(x^{2}+y^{2})\,{\it dx}\,{\it dy}
4\pi
\displaystyle\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\sqrt{4-x^{2}-y^{2}}\, {\it dy}\,{\it dx}
\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-x^{2}}}\cos (x^{2}+y^{2})\,{\it dy}\,{\it dx}
\dfrac{\pi}{4} \sin (1)
\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}e^{\sqrt{x^{2}+y^{2}}}\,{\it dx}\,{\it dy}
\iint\limits_{\kern-3ptR}e^{-(x^{2}+y^{2})}\,d\!A, R is the region in the first quadrant enclosed by the circles x^{2}+y^{2}=1 and x^{2}+y^{2}=4.
\dfrac{\pi(e^3-1)}{4e^4}
\displaystyle\iint\limits_{\kern-3ptR}\dfrac{y}{\sqrt{x^{2}+y^{2}}}\,d\!A, R is the region in the first quadrant inside the circle x^{2}+y^{2}=a^{2}.
\iint\limits_{\kern-3ptR}x\,d\!A, R is the region enclosed by the circle x^{2}+y^{2}=x.
\dfrac{\pi}8
\iint\limits_{\kern-3ptR}y^{2}\,d\!A, R is the region enclosed by the circle x^{2}+y^{2}=2y.
\iint\limits_{\kern-3ptR}\sqrt{x^{2}+y^{2}}\,d\!A, R is the region enclosed by r=3+\cos \theta .
21 \pi
\iint\limits_{\kern-3ptR}(x^{2}+y^{2})\,d\!A, R is the region enclosed by r=2(1+\sin \theta ).
\displaystyle\int_{-2}^{2}\int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}} }(x^{2}+y^{2})^2\,{\it dx}\,{\it dy}
\dfrac{64\pi}3
\displaystyle\int_{0}^{2}\int_{\sqrt{2x-x^{2}}}^{\sqrt{4x-x^{2}}}\,{\it dy}\,{\it dx}+\int_{2}^{4}\int_{0}^{\sqrt{ 4x-x^{2}}}\,{\it dy}\,{\it dx}
Applications and Extensions
In Problems 43–52, use double integrals in polar coordinates.
Volume Find the volume of the solid enclosed by the ellipsoid x^{2}+y^{2}+4z^{2}=4.
V = \dfrac{16\pi}3 cubic units
Volume Find the volume of the solid enclosed by the paraboloid x^{2}+y^{2}=az, a > 0, the xy-plane, and the cylinder x^{2}+y^{2}=a^{2}.
Volume Find the volume of the portion of the ball (a solid sphere) of radius 4 and center at the origin that lies above the region enclosed by the circle r=4\sin \theta , as shown in the figure below.
V = \dfrac{64\pi}3 - \dfrac{256}9 cubic units
Volume Find the volume of the solid cut from the ellipsoid x^{2}+y^{2}+4z^{2}=4 by the cylinder x^{2}+y^{2}=1.
Area Find the area enclosed by one loop of r^{2}=9\sin (2\theta) .
A = \dfrac92 square units
Area Find the area of the region that lies inside the circle r=4\cos \theta but outside the circle r=\cos \theta .
Area Find the area of the region in the first quadrant that lies inside the limaçon r=3+2\sin \theta but outside the circle r=4\sin \theta . See the figure.
A = 6 + \dfrac{3\pi}4 square units
Area Find the area of the region that lies inside the circle r=1 but outside the cardioid r=1+\cos \theta .
Area Find the area of the region that lies inside the cardioid r=1+\cos \theta but outside the circle r=\dfrac{1}{2}.
A = \dfrac{5\pi}6 + \dfrac{7 \sqrt{3}}8 square units
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Area Find the area of the region that lies inside the limaçon r=3-\cos \theta but outside the circle r=5\cos \theta .
In Problems 53 and 54, replace each iterated integral in polar coordinates with an iterated integral in rectangular coordinates. Do not find the integrals.
\displaystyle\int_{0}^{1/\sqrt{2}}\int_{0}^{\sin ^{-1}r}r\,d\theta \,dr
\int_0^{1/2} \int_{\sqrt{y-y^2}}^{\sqrt{1/2-y^2}} dx \, dy
\displaystyle\int_{0}^{1}\int_{\cos ^{-1}r}^{\pi /2}r^{2}\sin \theta \,d\theta \,dr
A tank in the shape of a hemisphere of radius 1 m is lying on its flat side. If the tank is filled with liquid to a depth of a meters, what is the volume of the liquid? (Hint: Position the hemisphere as shown in the figure.)
V = \pi a \!\left( 1 - \dfrac{a^2}{3} \right) cubic meters
Repeat Problem 55 if the tank is configured as in the figure, with its flat side at ground level and its rounded base under the ground.
Area
Volume Find the volume of the solid cut from the sphere x^{2}+y^{2}+z^{2}=a^{2} by the cylinder x^{2}+y^{2}=ay, a\gt 0.
Area Find the area enclosed by one leaf of the rose r=\sin (3\theta) .
A = \dfrac{\pi}{12} square units.
Challenge Problems
To find \int_{-\infty }^{\infty }e^{-x^{2}}\,{\it dx}, let I_{a}=\int_{0}^{a}e^{-x^{2}}\,{\it dx}.