극좌표 이중적분

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

$\int_{1}^{2}\,\int_{0}^{x}({x}^{2}+{y}^{2})dy\,dx=\int_{0}^{\frac{\pi }{4}}\,\int_{\sec \,\theta }^{2\,\sec \,\theta }{r}^{3}dr\,d\theta$

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

$\int_{2}^{3}\,\int_{0}^{x}\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}dy\,dx=\int_{0}^{\pi /4}\,\int_{2\,\sec \,\theta }^{3\sec \,\theta }r\,\cos \,\theta \,dr\,d\theta$

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

$\int_{0}^{1}\,\int_{{x}^{2}}^{x}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}dy\,dx=\int_{0}^{\pi /4}\,\int_{0}^{\tan \,\theta \,\sec \,\theta }dr\,d\theta$

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

$\int_{0}^{1}\,\int_{{x}^{2}}^{x}\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}dy\,dx=\int_{0}^{\pi /4}\,\int_{0}^{\tan \,\theta \,\sec \,\theta }r\,\sin \,\theta \,dr\,d\theta$

A radial function $f$ is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, $f(x,y)=g(r),$ where $r=\sqrt{{x}^{2}+{y}^{2}}.$ Show that if $f$ is a continuous radial function, then $\iint_{D}f(x,y)dA=({\theta }_{2}-{\theta }_{1})[G({R}_{2})-G({R}_{1})],$ where $G'(r)=rg(r)$ and $(x,y)\in D=\{(r,\theta )|{R}_{1}\le r\le {R}_{2},{\theta }_{1}\le \theta \le {\theta }_{2}\},$ with $0\le {R}_{1}<{R}_{2}$ and $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi .$

Let $f(x,y)=\frac{F'(r)}{r}$ be a continuous radial function defined on the annular region $D=\{(r,\theta )|{R}_{1}\le r\le {R}_{2},0\le \theta \le 2\pi \},$ where $r=\sqrt{{x}^{2}+{y}^{2}},$ $0<{R}_{1}<{R}_{2},$ and $F$ is a differentiable function. Show that $\iint_{D}f(x,y)dA=2\pi [F({R}_{2})-F({R}_{1})].$

Let $f$ be a continuous function that can be expressed in polar coordinates as a product of a function of $r$ only and a function of $\theta$ only; that is, $f(x,y)=h(\theta ),$ where $(x,y)\in D=\{(r,\theta )|{R}_{1}\le r\le {R}_{2},{\theta }_{1}\le \theta \le {\theta }_{2}\},$ with $0\le {R}_{1}<{R}_{2}$ and $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi .$ Show that $\iint_{D}f(x,y)dA=\frac{1}{2}\left({R}_{2}^{2}-{R}_{1}^{2}\right)[H({\theta }_{2})-H({\theta }_{1})],$ where $H$ is an antiderivative of $h.$

Let $f$ be a continuous function that can be expressed in polar coordinates as a product of a function of $r$ only and function of $\theta$ only; that is, $f(x,y)=g(r)h(\theta ),$ where $(x,y)\in D=\{(r,\theta )|{R}_{1}\le r\le {R}_{2},{\theta }_{1}\le \theta \le {\theta }_{2}\}$ with $0\le {R}_{1}<{R}_{2}$ and $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi .$ Show that $\iint_{D}f(x,y)dA=[G({R}_{2})-G({R}_{1})]\,[H({\theta }_{2})-H({\theta }_{1})],$ where $G(r)$ and $H(\theta )$ are antiderivatives of $rg(r)$ and $h(\theta )$, respectively.

A spherical cap is the region of a sphere that lies above or below a given plane.

a. Show that the volume of the spherical cap in the figure below is $\frac{1}{6}\pi h(3{a}^{2}+{h}^{2}).$ b. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is $h,$ show that the volume of the spherical segment in the figure below is $\frac{1}{6}\pi h(3{a}^{2}+3{b}^{2}+{h}^{2}).$

In statistics, the joint density for two independent, normally distributed events with a mean $\mu =0$ and a standard distribution $\sigma$ is defined by $p(x,y)=\frac{1}{2\pi {\sigma }^{2}}{e}^{-\frac{{x}^{2}+{y}^{2}}{2{\sigma }^{2}}}.$ Consider $(X,Y),$ the Cartesian coordinates of a ball in the resting position after it was released from a position on the z-axis toward the $xy$-plane. Assume that the coordinates of the ball are independently normally distributed with a mean $\mu =0$ and a standard deviation of $\sigma$ (in feet). The probability that the ball will stop no more than $a$ feet from the origin is given by $P[{X}^{2}+{Y}^{2}\le {a}^{2}]=\iint_{D}p(x,y)dy\,dx,$ where $D$ is the disk of radius a centered at the origin. Show that $P[{X}^{2}+{Y}^{2}\le {a}^{2}]=1-{e}^{-{a}^{2}/2{\sigma }^{2}}.$

The double improper integral $\int_{-\infty }^{\infty }\,\int_{-\infty }^{\infty }{e}^{-({x}^{2}+{y}^{2})/2}dy\,dx$ may be defined as the limit value of the double integrals $\iint_{{D}_{a}}{e}^{-({x}^{2}+{y}^{2})/2}dA$ over disks ${D}_{a}$ of radii a centered at the origin, as a increases without bound; that is, $\int_{-\infty }^{\infty }\,\int_{-\infty }^{\infty }{e}^{-({x}^{2}+{y}^{2})/2}dy\,dx=\lim_{a\to \infty }\iint_{{D}_{a}}{e}^{-({x}^{2}+{y}^{2})/2}dA.$

a. Use polar coordinates to show that $\int_{-\infty }^{\infty }\,\int_{-\infty }^{\infty }{e}^{-({x}^{2}+{y}^{2})/2}dy\,dx=2\pi .$ b. Show that $\int_{-\infty }^{\infty }{e}^{-{x}^{2}/2}dx=\sqrt{2\pi },$ by using the relation $\int_{-\infty }^{\infty }\,\int_{-\infty }^{\infty }{e}^{-({x}^{2}+{y}^{2})/2}dy\,dx=\left(\int_{-\infty }^{\infty }{e}^{-{x}^{2}/2}dx\right)\left(\int_{-\infty }^{\infty }{e}^{-{y}^{2}/2}dy\right).$