?A region \(R\) is shown. Decide whether to use polar coordinates or rectangular

A region \(R\) is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(x, y) d A\) as an iterated integral, where \(f\) is an arbitrary continuous function on \(R\). ________________ Equation Transcription: Text Transcription: R Double integral_R f(x,y)dA f R

A region \(R\) is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(x, y) d A\) as an iterated integral, where \(f\) is an arbitrary continuous function on \(R\). ________________ Equation Transcription: Text Transcription: R Double integral_R f(x,y)dA f R

A region \(R\) is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(x, y) d A\) as an iterated integral, where \(f\) is an arbitrary continuous function on \(R\). ________________ Equation Transcription: Text Transcription: R Double integral_R f(x,y)dA f R

A region \(R\) is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(x, y) d A\) as an iterated integral, where \(f\) is an arbitrary continuous function on \(R\). ________________ Equation Transcription: Text Transcription: R Double integral_R f(x,y)dA f R

A region \(R\) is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(x, y) d A\) as an iterated integral, where \(f\) is an arbitrary continuous function on \(R\). ________________ Equation Transcription: Text Transcription: R Double integral_R f(x,y)dA f R

1–6 A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_{R} f(\mathrm{x}, \mathrm{y}) \mathrm{d} \mathrm{A}\) as an iterated integral, where f is an arbitrary continuous function on R. Equation Transcription: ??(x, y)dA Text Transcription: Integral Integral_R f(x,y)dA

7-8 Sketch the region whose area is given by the integral and evaluate the integral. \(\int_{\pi / 4}^{3 \pi / 4} \int_{1}^{2} r d r d \theta\) Equation Transcription: ?? Text Transcription: Integral_pi/4^3pi/4 Integral_1^2 r dr d theta Image text transcription: Sketch the region whose area is given by the integral and evaluate the integral. ?/43/4 ?12 r dr d

7-8 Sketch the region whose area is given by the integral and evaluate the integral. \(\int _{\pi / 2}^\pi \int_{0}^{2 \sin \theta} r \ d r \ d \theta\) Equation Transcription: ? ? Text Transcription: Integral_pi/2^pi Integral_0^2 sin theta r dr d theta

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{D} x^{2} y d A\)=, where D is the top half of the disk with center the origin and radius 5 Equation Transcription: ?? Text Transcription: Integral Integral_D x^2 y dA

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{R}(2 x-y) d A\), where R is the region in the first quadrant enclosed by the circle \(x^{2}+y^{2}=4\) and the lines \(x=0\) and \(y=x\) Equation Transcription: ?? Text Transcription: Integral Integral_R (2x-y)dA x^2+y^2=4 x=0 y=x

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{R} \sin \left(x^{2}+y^{2}\right) d A\), where R is the region in the first quadrant between the circles with center the origin and radii 1 and 3 Equation Transcription: ?? Text Transcription: Integral Integral_R sin(x^2+y^2)dA

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A\), where R is the region that lies between the circles \(x^{2}+y^{2}=a^{2}\) and \(x^{2}+y^{2}=b^{2}\) with \(0<a<b\) Equation Transcription: ?? Text Transcription: Integral Integral_R y^2/x^2+y^2 dA x^2+y^2=a^2 x^2+y^2=b^2 0 < a < b Image text transcription: Evaluate the given integral by changing to polar coordinates. ?R y2x2+y2dA, where R is the region that lies between the circles x2+y2=a2 and x2+y2=b2 with 0<a<b

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{R} e^{-x^{2-y^{2}}} d A\) , where D is the region bounded by the semicircle \(x=\sqrt{4-y^{2}}\) and the -axis Equation Transcription: ?? Text Transcription: Integral Integral_R e^-x^2-y^2 x=sqrt 4-y^2

9-16 Evaluate the given integral by changing to polar coordinates. \(Integral Integral_R\), where D is the disk with center the origin and radius 2 Equation Transcription: ?? Text Transcription: Integral Integral_D cos sqrt x^2+y^2 dA

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{D} \arctan (y / x) d A\), where \(R=\left\{(x, y) \mid 1 \leqslant x^{2}+y^{2} \leqslant 4,0 \leqslant y \leqslant x\right\}\) Equation Transcription: ?? R = {(x,y) |1 ? x2 + y2 ? 4,0 ? y ? x} Text Transcription: Integral Integral_D arctan (y/x)dA R={(x,y)|1 leqslant x^2 + y^2 leqslant 4, 0 leqslant y leqslant x} Image text transcription: Evaluate the given integral by changing to polar coordinates. ?R arctan?(y/x)dA, where R=(x,y)?1 ? x2+y2 ? 4,0 ? y ?x

9-16 Evaluate the given integral by changing to polar coordinates. \(\iint_{D} x \ d A\), where D is the region in the first quadrant that lies between the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}=2 x\) Equation Transcription: ?? Text Transcription: Integral Integral_D x dA x^2+y^2=4 x^2+y^2=2x Image text transcription: Evaluate the given integral by changing to polar coordinates. ?D xdA, where D is the region in the first quadrant that lies between the circles x2+y2=4 and x2+y2=2x

\(17-22\) Use a double integral to find the area of the region D. Equation Transcription: Text Transcription: 17-22

\(17-22\) Use a double integral to find the area of the region D. Equation Transcription: Text Transcription: 17-22

\(17-22\) Use a double integral to find the area of the region D. Equation Transcription: Text Transcription: 17-22

\(17-22\) Use a double integral to find the area of the region D. Equation Transcription: Text Transcription: 17-22

\(17-22\) Use a double integral to find the area of the region D. D is the loop of the rose \(r=\sin 3 \theta\) in the first quadrant. Equation Transcription: r = sin Text Transcription: 17-22 r=sin 3 theta Image text transcription: Use a double integral to find the area of the region D. D is the loop of the rose r = sin3 in the first quadrant.

\(17-22\) Use a double integral to find the area of the region D. D is the region inside the circle \((x-2)^{2}+y^{2}=1\) and outside the circle \(x^{2}+y^{2}=1\). Equation Transcription: (x - 2 )2 + y2 = 1 x2 + y2 = 1 Text Transcription: 17-22 (x - 2 )^2 + y^2 = 1 x^2 + y^2 = 1 Image text transcription: Use a double integral to find the area of the region D. D is the region inside the circle (x - 2 )2 + y2 = 1 and outside the circle x2 + y2 = 1.

\(23–24\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the surface and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. Equation Transcription: Text Transcription: 23–24

\(23–24\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the surface and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. Equation Transcription: Text Transcription: 23–24

\(25–28\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the graph of the given function and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. \(f(x, y) = y\) Equation Transcription: Text Transcription: 25–28 f(x, y) = y

\(25–28\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the graph of the given function and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. \(f(x, y)=x y^{2}\) Equation Transcription: f(x, y) = xy2 Text Transcription: 25–28 f(x, y) = xy^2

\(25–28\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the graph of the given function and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. \(f(x, y) = x\) Equation Transcription: Text Transcription: 25–28 f(x, y) = x

29-37 Use polar coordinates to find the volume of the given solid. Under the paraboloid \(z=x^{2}+y^{2}\) and above the disk \(x^{2}+y^{2} \leqslant 25\) Equation Transcription: ? 25 Text Transcription: z=x^2+y^2 x^2+y^2 leqslant 25 Image text transcription: Use polar coordinates to find the volume of the given solid. Under the paraboloid z=x2+y2and above the disk x2+y2 ? 25

29-37 Use polar coordinates to find the volume of the given solid. Below the cone \(z=\sqrt{x^{2}+y^{2}}\)and above the ring \(1 \leqslant x^{2}+y^{2} \leqslant 4\) Equation Transcription: 1 ? x2 + y2 ? 4 Text Transcription: z=sqrt x^2 + y^2 1 leqslant x^2 + y^2 leqslant 4

29-37 Use polar coordinates to find the volume of the given solid. Below the plane \(2 x+y+z=4\)and above the disk \(x^{2}+y^{2} \leqslant 1\) Equation Transcription: ? 1 Text Transcription: 2x+y+z=4 x^2+y^2 leqslant 4

29-37 Use polar coordinates to find the volume of the given solid. Inside the sphere \(x^{2}+y^{2}+z^{2}=16\) and outside the cylinder \(x^{2}+y^{2}=4\) Equation Transcription: Text Transcription: x^2+y^2+z^2=16 x^2+y^2=4 Image text transcription: Use polar coordinates to find the volume of the given solid. Inside the sphere x2+y2+z2=16 and outside the cylinder x2+y2=4

\(29-37\) Use polar coordinates to find the volume of the given solid. A sphere of radius a Equation Transcription: Text Transcription: 29-37

\(29-37\) Use polar coordinates to find the volume of the given solid. Bounded by the paraboloid \(z=1+2 x^{2}+2 y^{2}\) and the plane \(z=7\) in the first octant Equation Transcription: Text Transcription: 29-37 z=1+2x^2+2y^2 z=7 Image text transcription: Use polar coordinates to find the volume of the given solid. Bounded by the paraboloid z=1+2x2+2y2 and the plane z=7 in the first octant

\(29-37\) Use polar coordinates to find the volume of the given solid. Above the cone \(z=\sqrt{x^{2}+y^{2}}\) and below the sphere \(x^{2}+y^{2}+z^{2}=1\) Equation Transcription: Text Transcription: 29-37 z=sqrt x^2+y^2 x^2+y^2+z^2=1 Image text transcription: Use polar coordinates to find the volume of the given solid. Above the cone z=x2+y2 and below the sphere x2+y2+z2=1

\(29-37\) Use polar coordinates to find the volume of the given solid. Bounded by the paraboloids \(z=6-x^{2}-y^{2}\) and \(z=2 x^{2}+2 y^{2}\) Equation Transcription: Text Transcription: 29-37 z=6-x^2-y^2 z=2x^2+2y^2 Image text transcription: Use polar coordinates to find the volume of the given solid. Bounded by the paraboloids z=6?x2?y2 and z=2x2+2y2

\(29-37\) Use polar coordinates to find the volume of the given solid. Inside both the cylinder \(x^{2}+y^{2}=4\) and the ellipsoid \(4 x^{2}+4 y^{2}+z^{2}=64\) Equation Transcription: Text Transcription: 29-37 x^2+y^2=4 4x^2+4y^2+z^2=64 Image text transcription: Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2+y2=4 and the ellipsoid 4x2+4y2+z2=64

(a) A cylindrical drill with radius \(r_{1}\) is used to bore a hole through the center of a sphere of radius \(r_{2}\). Find the volume of the ring-shaped solid that remains. (b) Express the volume in part (a) in terms of the height h of the ring. Notice that the volume depends only on h, not on \(r_{1}\) or \(r_{2}\). Equation Transcription: Text Transcription: r_1 r_2

39-42 Evaluate the iterated integral by converting to polar coordinates. \(\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} e^{-x^{2-y^{2}}} d y \ d x\) Equation Transcription: ? ? Text Transcription: Integral_0^2 Integral_0^sqrt 4-x^2 e^-x^2-y2 dy dx

39-42 Evaluate the iterated integral by converting to polar coordinates. \(\int_{0}^{a} \int_{-\sqrt{a^{2}-y^{3}}}^{\sqrt{a-x^{2}}}(2 x+y) d x d y\) Equation Transcription: ? ? Text Transcription: Integral_0^a Integral_-sqer a^2-y^3

39–42 Evaluate the iterated integral by converting to polar coordinates. \(\int_{0}^{1 / 2} \int_{\sqrt{5 y}}^{\sqrt{1-y^{2}}} x y^{2} \ d x \ d y\) Equation Transcription: ? ? Text Transcription: Integral_0^1/2 Integral_sqrt 5y^sqrt 1-y^2 xy^2 dx dy

39–42 Evaluate the iterated integral by converting to polar coordinates. \(\int_{0}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \sqrt{x^{2}+y^{2}} \ d y \ d x\) Equation Transcription: ? ? Text Transcription: Integral_0^2 Integral_0^sqrt 2x-x^2 sqrt x^2+y^2 dy dx

43-44 Express the double integral in terms of a single integral with respect to r. Then use a calculator (or computer) to evaluate the integral correct to four decimal places. \(\iint_{D} e^{\left(x^{2}+y^{2}\right)^{2}} d A\), where D is the disk with center the origin and radius 1. Equation Transcription: ?? Text Transcription: Integral Integral_D e^(x^2+y^2)^2 dA

43-44 Express the double integral in terms of a single integral with respect to r. Then use a calculator (or computer) to evaluate the integral correct to four decimal places. \(\iint_{D} x y \sqrt{1+x^{2}+y^{2}} \ d A\), where is the portion of the disk \(x^{2}+y^{2} \leq 1\) that lies in the first quadrant. Equation Transcription: ?? Text Transcription: Integral Integral_D xy sqrt 1+x^2+y^2 dA x^2+y^2 leq 1

A swimming pool is circular with a 40-ft diameter. The depth is constant along east-west lines and increases linearly from \(2 f t\) at the south end to \(7 f t\) at the north end. Find the volume of water in the pool. Equation Transcription: Text Transcription: 2ft 7ft

An agricultural sprinkler distributes water in a circular pattern of radius \(100 \mathrm{ft}\). It supplies water to a depth of \(e^{-r}\) feet per hour at a distance of r feet from the sprinkler. (a) If \(0<R \leqslant 100\), what is the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler? (b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius R. Equation Transcription: ? 100 Text Transcription: 100ft e^-r 0 < R leqslant 100

Find the average value of the function \(f(x, y)=1 / \sqrt{x^{2}+y^{2}}\)on the annular region \(a^{2} \leqslant x^{2}+y^{2} \leqslant b^{2}\), where \(0<a<b\). Equation Transcription: ? ? Text Transcription: f(x,y)=1/sqrt x^2+y^2 a^2 leqslant x^2 +y^2 leqslant b^2 0 < a < b

Let \(D\) be the disk with center the origin and radius \(a\). What is the average distance from points in \(D\) to the origin? Equation Transcription: Text Transcription: D a

Use polar coordinates to combine the sum \(\int_{1 / \sqrt{2}}^{1} \int_{\sqrt{1-x^{2}}}^{x} x y \ d y \ d x+\int_{1}^{\sqrt{2}} \int_{0}^{x} x y \ d y \ d x+\int_{\sqrt{2}}^{2} \int_{0}^{\sqrt{4-x^{2}}} x y \ d y \ d x\) into one double integral. Then evaluate the double integral. Equation Transcription: ? ?? ? ? ? Text Transcription:

(a) We define the improper integral (over the entire plane \(R^{2}\) ) \(I=\iint_{R^{2}} e^{-\left(x^{2}+y^{2}\right)} d A=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d y \ d x=\lim _{a \rightarrow \infty} \iint_{D_{a}} e^{-\left(x^{2}+y^{2}\right)} d A\) where \(D_{a}\) is the disk with radius \(a\) and center the origin. Show that \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d A=\pi\) (b) An equivalent definition of the improper integral in part (a) is \(\iint_{A^{2}} e^{-\left(x^{2}+y^{2}\right)} d A=\lim _{a \rightarrow \infty} \iint_{S_{a}} e^{-\left(x^{2}+y^{2}\right)} d A\) where \(S_{a}\) is the square with vertices \((\pm a, \pm a)\). Use this to show that \(\int_{-\infty}^{\infty} e^{-x^{2}} d x \int_{-\infty}^{\infty} e^{-y^{2}} d y=\pi\) (c) Deduce that \(\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}\) (d) By making the change of variable \(t=\sqrt{2} x\), show that \(\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}\) (This is a fundamental result for probability and statistics.) Equation Transcription: ?? ?? ?? ?? ?? ?? ?? ? ? Text Transcription: R^2 I=Integral Integral_R^2 e^-(x^2+y^2) dA=Integral_-infinity^infinity Integral_-infinity^infinity e^-(x^2+y^2) dy dx = lim_a rightarrow infinity e^-(x^2+y^2) dA D_a a Integral_-infinity^infinity Integral_-infinity^infinity e^-(x^2+y^2) dA=pi Integral Integral_A^2 e^-(x^2+y^2) dA=lim_a rightarrow infinity Integral Integral_S_a e^-(x^2+y^2) dA S_a (pm a, pm a) Integral_-infinity^infinity e^-x^2 dx Integral_-infinity^infinity e^-y^2 dy=pi Integral_-infinity^infinity e^-x^2 dx=sqrt pi t=sqrt 2x Integral_-infinity^infinity e^-x^2/2 dx=sqrt 2 pi

Use the result of Exercise to evaluate the following integrals. (a) \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (b) \(\int_{0}^{\infty} \sqrt{x} e^{-x} d x\) Equation Transcription: ? ? Text Transcription: Integral_0^infinity x^2 e^-x^2 dx Integral_0^infinity sqrt xe ^-x dx

\(25–28\) (a) Set up an iterated integral in polar coordinates for the volume of the solid under the graph of the given function and above the region D. (b) Evaluate the iterated integral to find the volume of the solid. \(f(x, y) = 1\) Equation Transcription: Text Transcription: 25–28 f(x, y) = 1