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Convert the function \(f(x,y) =\sqrt{4x^{2}+4y^{2}}\) to polar coordinates.

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In polar coordinates, the differential \({\it dA}\) of area is \(dA\) = _________.

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True or False Geometrically, if \(f(x,y)\geq 0\) and if \(f\) is continuous over a closed, bounded region \(R\), then the double integral \( \iint\limits_{\kern-3ptR}f(r\cos\theta, r \sin \theta )\,{dr d\theta}\) represents the volume of the solid under the surface \(z=f(x,y)\) and above the region \(R\) in the \(xy\)-plane.

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True or False \(\iint\limits_{\kern-3ptR}r\,dr\,d\theta\) equals the area \(A\) of the closed, bounded region \(R\).

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\(\iint\limits_{\kern-3ptR} ( x+y) \, {\it dA},\) \(R\) is enclosed by \(x^{2}+y^{2}=9\), \(x\geq 0\), and \(y\geq 0\).

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\(\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \, {\it dA},\) \(R\) is enclosed by \(y=\sqrt{1-x^{2}}\) and the lines \(y=x\) and \(y=-x.\)

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\(\iint\limits_{\kern-3ptR}x\, {\it dA},\) \(R\) is enclosed by \(x=\sqrt{4-y^{2}},\) the positive \(x\)-axis, and the line \(y=x.\)

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\(\iint\limits_{\kern-3ptR}y\, {\it dA},\) \(R\) is enclosed by \(y=\sqrt{16-x^{2}},\) in the first quadrant.

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\(\iint\limits_{\kern-3ptR}( x+y^{2}) \, {\it dA},\) \(R\) is enclosed by \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\).

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\(\iint\limits_{\kern-3ptR}\sqrt{x^{2}+y^{2}}\, {\it dA},\) \(R\) is enclosed by \(x^{2}+y^{2}=\dfrac{1}{4}\) and \(x^{2}+y^{2}=5\).

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\(\displaystyle\int_{0}^{\pi /4}\int_{0}^{4\sin \theta }r\cos \theta \,{\it dr}\,d\theta \)

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\(\displaystyle\int_{0}^{5\pi /3}\int_{0}^{2\cos \theta }r\sin \theta \,{\it dr}\,d\theta \)

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\(\displaystyle\int_{0}^{\pi /2}\int_{0}^{\pi /3}r\,\,d\theta\, {\it dr}\)

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\(\displaystyle\int_{0}^{\pi /3}\int_{0}^{\sin \theta }r\,{\it dr}\,d\theta \)

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\(\displaystyle\int_{0}^{\pi /4}\int_{0}^{\cos \theta }r\,{\it dr}\,d\theta \)

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\(\displaystyle\int_{0}^{4}\int_{-\pi /4}^{\pi /4}r\cos \theta \,d\theta\, {\it dr}\)

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\(\displaystyle\int_{0}^{2}\int_{-\pi /3}^{5\pi /3}r\sin \theta \,d\theta\, {\it dr} \)

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\(\displaystyle\int_{0}^{\pi}\int_{0}^{\pi /4}r\,\,d\theta\, {\it dr}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}3\sin \theta \,{\it dA} ,\quad\)\(0\leq r\leq 2,\) \(0\leq \theta \leq \dfrac{\pi }{2}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}4\cos \theta \,{\it dA} ,\quad\) \(0\leq r\leq 3,\) \(0\leq \theta \leq \dfrac{\pi }{2}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}2r \sin \theta \,{\it dA} ,\quad\) \(0\leq r\leq 1,\) \(0\leq \theta \leq \dfrac{\pi }{2}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}3r \cos \theta \,{\it dA} ,\quad\) \(0\leq r\leq 2\), \(0\leq \theta \leq \dfrac{\pi }{4}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}3r^{2}\sin \theta \,{\it dA} ,\quad\) \(0\leq r\leq 2,\) \(-\dfrac{\pi }{2}\leq \theta \leq \dfrac{\pi }{2}\)

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\(\displaystyle\iint\limits_{\kern-3ptR}2r^{2}\cos \theta \,{\it dA} ,\quad\) \(0\leq r\leq 4,\) \(- \dfrac{\pi }{2}\leq \theta \leq \dfrac{\pi }{2}\)

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\(\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-x^{2}}}\,{\it dy}\,{\it dx}\)

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\(\displaystyle\int_{0}^{3}\int_{0}^{y}\sqrt{x^{2}+y^{2}}\,{\it dx}\,{\it dy}\)

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\(\displaystyle\int_{-2}^{2}\int_{0}^{\sqrt{4-y^{2}}}(x^{2}+y^{2})\,{\it dx}\,{\it dy}\)

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\(\displaystyle\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\sqrt{4-x^{2}-y^{2}}\, {\it dy}\,{\it dx}\)

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\(\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-x^{2}}}\cos (x^{2}+y^{2})\,{\it dy}\,{\it dx}\)

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\(\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}e^{\sqrt{x^{2}+y^{2}}}\,{\it dx}\,{\it dy}\)

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\(\iint\limits_{\kern-3ptR}e^{-(x^{2}+y^{2})}\,d\!A,\) \(R\) is the region in the first quadrant enclosed by the circles \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\).

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\(\displaystyle\iint\limits_{\kern-3ptR}\dfrac{y}{\sqrt{x^{2}+y^{2}}}\,d\!A,\) \(R\) is the region in the first quadrant inside the circle \(x^{2}+y^{2}=a^{2}.\)

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\(\iint\limits_{\kern-3ptR}x\,d\!A,\) \(R\) is the region enclosed by the circle \(x^{2}+y^{2}=x\).

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\(\iint\limits_{\kern-3ptR}y^{2}\,d\!A,\) \(R\) is the region enclosed by the circle \(x^{2}+y^{2}=2y\).

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\(\iint\limits_{\kern-3ptR}\sqrt{x^{2}+y^{2}}\,d\!A,\) \(R\) is the region enclosed by \(r=3+\cos \theta \).

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\(\iint\limits_{\kern-3ptR}(x^{2}+y^{2})\,d\!A,\) \(R\) is the region enclosed by \(r=2(1+\sin \theta )\).

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\(\displaystyle\int_{-2}^{2}\int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}} }(x^{2}+y^{2})^2\,{\it dx}\,{\it dy}\)

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\(\displaystyle\int_{0}^{2}\int_{\sqrt{2x-x^{2}}}^{\sqrt{4x-x^{2}}}\,{\it dy}\,{\it dx}+\int_{2}^{4}\int_{0}^{\sqrt{ 4x-x^{2}}}\,{\it dy}\,{\it dx}\)

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Volume Find the volume of the solid enclosed by the ellipsoid \(x^{2}+y^{2}+4z^{2}=4\).

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Volume Find the volume of the solid enclosed by the paraboloid \(x^{2}+y^{2}=az\), \(a > 0\), the \(xy\)-plane, and the cylinder \( x^{2}+y^{2}=a^{2}\).

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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.

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Volume Find the volume of the solid cut from the ellipsoid \( x^{2}+y^{2}+4z^{2}=4\) by the cylinder \(x^{2}+y^{2}=1\).

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Area Find the area enclosed by one loop of \(r^{2}=9\sin (2\theta) \).

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Area Find the area of the region that lies inside the circle \( r=4\cos \theta \) but outside the circle \(r=\cos \theta \).

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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.

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Area Find the area of the region that lies inside the circle \(r=1\) but outside the cardioid \(r=1+\cos \theta \).

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Area Find the area of the region that lies inside the cardioid \(r=1+\cos \theta \) but outside the circle \(r=\dfrac{1}{2}\).

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Area Find the area of the region that lies inside the limaçon \(r=3-\cos \theta \) but outside the circle \(r=5\cos \theta \).

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\(\displaystyle\int_{0}^{1/\sqrt{2}}\int_{0}^{\sin ^{-1}r}r\,d\theta \,dr\)

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\(\displaystyle\int_{0}^{1}\int_{\cos ^{-1}r}^{\pi /2}r^{2}\sin \theta \,d\theta \,dr\)

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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.)

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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.

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Area

1. Use technology to graph \(r=\theta ,\) \( 0\leq \theta \leq 2\pi .\)
2. Find the area of the region enclosed by the graph of \(r=\theta\) and the positive \(x\)-axis.

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Volume Find the volume of the solid cut from the sphere \( x^{2}+y^{2}+z^{2}=a^{2}\) by the cylinder \(x^{2}+y^{2}=ay\), \(a\gt 0\).

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Area Find the area enclosed by one leaf of the rose \(r=\sin (3\theta) \).

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1. Express \(f(x,y) =\dfrac{{1}}{x^{2} \sqrt{x^{2}+y^{2}}}\,,\) using polar coordinates.
2. Find \(\displaystyle\int_{1}^{2}{\int_{0}^{1}} \dfrac{1}{x^{2}\sqrt{x^{2}+y^{2}}}\, {\it dy}\,{\it dx}.\)

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To find \(\int_{-\infty }^{\infty }e^{-x^{2}}\,{\it dx}\), let \( I_{a}=\int_{0}^{a}e^{-x^{2}}\,{\it dx}\).

1. Show that: \[ I_{a}^{2}=\int_{0}^{a}e^{-x^{2}}\,{\it dx}\int_{0}^{a}e^{-y^{2}} {\it dy}=\int_{0}^{a} \int_{0}^{a}e^{-(x^{2}+y^{2})}\,{\it dx}\,{\it dy} \]
2. Let \(J_{a}=\iint\limits_{\kern-3ptR}e^{-(x^{2}+y^{2})}\,{\it dA}\), where \(R\) is the quarter circle \(0\leq \theta \leq \dfrac{\pi }{2}\), \( 0\leq r\leq a\) . Show that \[ \big\vert I_{a}^{2}-J_{a}\big\vert \leq \dfrac{(4-\pi )a^{2}}{4}e^{-a^{2}} \]
3. Find \(J_{a}\).
4. Show that \[ \lim_{a\rightarrow \infty }J_{a}=\dfrac{\pi }{4}\qquad \hbox{and}\qquad \lim_{a\rightarrow \infty }\,\big\vert \,I_{a}^{2}-J_{a}\big\vert =0 \]
5. Show that \(\int_{-\infty }^{\infty }e^{-x^{2}}\,{\it dx}=\sqrt{\pi }\). (This integral is of special importance in statistics.)