## 11 номер – Д 2239
#учеба #семестр_1 #высшая_математика


### Пример:

$\int_{0}^{\ln2} xe^{ -x } \, dx$

### Решение:

$I=\int_{0}^{\ln2} xe^{-x}dx$

$I=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x;\ du=dx$
$dv=e^{-x}dx;\ v=-e^{-x}$

$I=.x(-e^{-x})|_{0}^{\ln2}-\int_{0}^{\ln2}(-e^{-x})dx=.-xe^{-x}|_{0}^{\ln2}+\int_{0}^{\ln2}e^{-x}dx$

$\int e^{-x}dx=-e^{-x}$
$I=.-xe^{-x}|_{0}^{\ln2}+.-e^{-x}|_{0}^{\ln2}$

$e^{-\ln2}=\dfrac12$
$I=-(\ln2)\cdot\dfrac12- \dfrac12- (0\cdot1-1)=-\dfrac{\ln2}{2}-\dfrac12+1=\dfrac12-\dfrac{\ln2}{2}$

## 13 номер – Д 2241

### Пример:

$\int_{0}^{2\pi} x^{2}\cos x \, dx$

### Решение:

$I=\int_{0}^{2\pi} x^{2}\cos x\,dx$

$I=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x^{2};\ du=2x\,dx$
$dv=\cos x\,dx;\ v=\sin x$

$I=.x^{2}\sin x|_{0}^{2\pi}-\int_{0}^{2\pi}2x\sin x\,dx=-2\int_{0}^{2\pi}x\sin x\,dx$

$J=\int_{0}^{2\pi}x\sin x\,dx$
$J=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x;\ du=dx$
$dv=\sin x\,dx;\ v=-\cos x$

$J=.-x\cos x|_{0}^{2\pi}+\int_{0}^{2\pi}\cos x\,dx=.-x\cos x|_{0}^{2\pi}+.\sin x|_{0}^{2\pi}$
$J=-2\pi\cdot1+0-0=-2\pi$

$I=-2J=-2(-2\pi)=4\pi$

## 15 номер – Д 2244

### Пример:

$\int_{0}^{\sqrt{ 3 }} x\ \text{arccot} x\, dx$

### Решение:

$I=\int_{0}^{\sqrt3}x\,\text{arccot}\,x\,dx$

$I=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=\text{arccot}\,x;\ du=-\dfrac{1}{1+x^{2}}dx$
$dv=x\,dx;\ v=\dfrac{x^{2}}{2}$

$I=.\dfrac{x^{2}}{2}\text{arccot}\,x|_{0}^{\sqrt3}-\int_{0}^{\sqrt3}\dfrac{x^{2}}{2}(-\dfrac{1}{1+x^{2}})dx=.\dfrac{x^{2}}{2}\text{arccot}\,x|_{0}^{\sqrt3}+\dfrac12\int_{0}^{\sqrt3}\dfrac{x^{2}}{1+x^{2}}dx$

$\dfrac{x^{2}}{1+x^{2}}=1-\dfrac{1}{1+x^{2}}$
$I=.\dfrac{x^{2}}{2}\text{arccot}\,x|_{0}^{\sqrt3}+\dfrac12\int_{0}^{\sqrt3}(1-\dfrac{1}{1+x^{2}})dx$
$\int \dfrac{dx}{1+x^{2}}=\arctan x$
$I=.\dfrac{x^{2}}{2}\text{arccot}\,x|_{0}^{\sqrt3}+\dfrac12.(x-\arctan x)|_{0}^{\sqrt3}$

$\text{arccot}\sqrt3=\dfrac{\pi}{6};\ \arctan\sqrt3=\dfrac{\pi}{3}$
$I=\dfrac{3}{2}\cdot\dfrac{\pi}{6}+\dfrac12(\sqrt3-\dfrac{\pi}{3})=\dfrac{\pi}{4}+\dfrac{\sqrt3}{2}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}+\dfrac{\pi}{12}$

## 17 номер – Д 2246

### Пример:

$\int_{0}^{a} x^{2}\sqrt{ a^{2}-x^{2} } \, dx$

### Решение:

$a>0$
$I=\int_{0}^{a} x^{2}\sqrt{a^{2}-x^{2}}\,dx$

$x=a\sin t;\ dx=a\cos t\,dt;\ \sqrt{a^{2}-x^{2}}=a\cos t$
$x=0\implies t=0;\ x=a\implies t=\dfrac{\pi}{2}$

$I=\int_{0}^{\pi/2}(a^{2}\sin^{2}t)\cdot(a\cos t)\cdot(a\cos t)\,dt=a^{4}\int_{0}^{\pi/2}\sin^{2}t\cos^{2}t\,dt$

$\sin^{2}t\cos^{2}t=\dfrac14\sin^{2}2t;\ \sin^{2}2t=\dfrac{1-\cos4t}{2}$
$I=a^{4}\int_{0}^{\pi/2}\dfrac18(1-\cos4t)\,dt=\dfrac{a^{4}}{8}(t-\dfrac{\sin4t}{4})\Big|_{0}^{\pi/2}$

$I=\dfrac{a^{4}}{8}(\dfrac{\pi}{2}-0)=\dfrac{\pi a^{4}}{16}$

## 19 номер – Д 2248

### Пример:

$\int_{0}^{\ln 2} \sqrt{ e^{ x }-1 } \, dx$

### Решение:

$I=\int_{0}^{\ln2}\sqrt{e^{x}-1}\,dx$

$t=e^{x};\ dt=e^{x}dx;\ dx=\dfrac{dt}{t}$
$x=0\implies t=1;\ x=\ln2\implies t=2$

$I=\int_{1}^{2}\dfrac{\sqrt{t-1}}{t}\,dt$

$u=\sqrt{t-1};\ t=u^{2}+1;\ dt=2u\,du$
$t=1\implies u=0;\ t=2\implies u=1$

$I=\int_{0}^{1}\dfrac{u}{u^{2}+1}\cdot2u\,du=2\int_{0}^{1}\dfrac{u^{2}}{u^{2}+1}\,du$
$\dfrac{u^{2}}{u^{2}+1}=1-\dfrac{1}{u^{2}+1}$

$I=2\int_{0}^{1}(1-\dfrac{1}{u^{2}+1})du=2(u-\arctan u)\Big|_{0}^{1}$
$I=2(1-\dfrac{\pi}{4})=2-\dfrac{\pi}{2}$

## 21 номер – Д 2269

### Пример:

$\int_{-1}^{1} \dfrac{xdx}{x^{2}+x+1}$

### Решение:

$I=\int_{-1}^{1}\dfrac{x}{x^{2}+x+1}dx$

$x=\dfrac12(2x+1)-\dfrac12$
$I=\dfrac12\int_{-1}^{1}\dfrac{2x+1}{x^{2}+x+1}dx-\dfrac12\int_{-1}^{1}\dfrac{dx}{x^{2}+x+1}$

$I_1=\dfrac12\int_{-1}^{1}\dfrac{2x+1}{x^{2}+x+1}dx=\dfrac12.\ln(x^{2}+x+1)|_{-1}^{1}=\dfrac12(\ln3-\ln1)=\dfrac12\ln3$

$x^{2}+x+1=(x+\dfrac12)^{2}+\dfrac34$
$I_2=-\dfrac12\int_{-1}^{1}\dfrac{dx}{(x+\dfrac12)^{2}+(\dfrac{\sqrt3}{2})^{2}}$

$\int\dfrac{dx}{(x-a)^{2}+b^{2}}=\dfrac{1}{b}\arctan\dfrac{x-a}{b}$
$I_2=-\dfrac12\cdot\dfrac{2}{\sqrt3}.\arctan(\dfrac{x+\frac12}{\frac{\sqrt3}{2}})|_{-1}^{1}=-\dfrac{1}{\sqrt3}.\arctan(\dfrac{2x+1}{\sqrt3})|_{-1}^{1}$

$\arctan\dfrac{3}{\sqrt3}=\arctan\sqrt3=\dfrac{\pi}{3};\ \arctan\dfrac{-1}{\sqrt3}=-\dfrac{\pi}{6}$
$I_2=-\dfrac{1}{\sqrt3}(\dfrac{\pi}{3}-(-\dfrac{\pi}{6}))=-\dfrac{1}{\sqrt3}\cdot\dfrac{\pi}{2}=-\dfrac{\pi}{2\sqrt3}$

$I=I_1+I_2=\dfrac12\ln3-\dfrac{\pi}{2\sqrt3}$

## 23 номер – Д 2271

### Пример:

$\int_{1}^{9} x\sqrt[ 3 ]{ 1-x } \, dx$

### Решение:

$I=\int_{1}^{9}x\sqrt[3]{1-x}\,dx$

$t=1-x;\ dt=-dx;\ x=1-t$
$x=1; t=0;\ x=9; t=-8$

$I=\int_{0}^{-8}(1-t)t^{\frac13}(-dt)=\int_{-8}^{0}(1-t)t^{\frac13}dt=\int_{-8}^{0}(t^{\frac13}-t^{\frac43})dt$

$I=.(\dfrac{3}{4}t^{\frac43}-\dfrac{3}{7}t^{\frac73})|_{-8}^{0}=-(\dfrac{3}{4}(-8)^{\frac43}-\dfrac{3}{7}(-8)^{\frac73})$

$\sqrt[3]{-8}=-2;\ (-8)^{\frac43}=(\sqrt[3]{-8})^{4}=(-2)^{4}=16;\ (-8)^{\frac73}=(\sqrt[3]{-8})^{7}=(-2)^{7}=-128$
$I=-(\dfrac{3}{4}\cdot16-\dfrac{3}{7}\cdot(-128))=-(12+\dfrac{384}{7})=-\dfrac{468}{7}$

## 25 номер – Д 2273

### Пример:

$\int_{0}^{1} x^{15} \, \sqrt{ 1+3x^{8} } \, dx$

### Решение:

$I=\int_{0}^{1}x^{15}\sqrt{1+3x^{8}}dx$

$u=1+3x^{8};\ du=24x^{7}dx;\ x^{7}dx=\dfrac{1}{24}du$
$x^{15}dx=x^{8}\cdot x^{7}dx$

$x=0;\ u=1$
$x=1;\ u=4$

$x^{8}=\dfrac{u-1}{3}$
$I=\int_{1}^{4}\dfrac{u-1}{3}\cdot\sqrt{u}\cdot\dfrac{1}{24}du=\dfrac{1}{72}\int_{1}^{4}(u-1)u^{\frac12}du=\dfrac{1}{72}\int_{1}^{4}(u^{\frac32}-u^{\frac12})du$

$I=\dfrac{1}{72}(\dfrac{2}{5}u^{\frac52}-\dfrac{2}{3}u^{\frac32})\Big|_{1}^{4}=\dfrac{1}{72}(\dfrac{2}{5}(4^{\frac52}-1)-\dfrac{2}{3}(4^{\frac32}-1))$
$4^{\frac52}=32;\ 4^{\frac32}=8$
$I=\dfrac{1}{72}(\dfrac{2}{5}\cdot31-\dfrac{2}{3}\cdot7)=\dfrac{1}{72}(\dfrac{62}{5}-\dfrac{14}{3})=\dfrac{1}{72}\cdot\dfrac{116}{15}=\dfrac{29}{270}$

## 27 номер – Д 2275

### Пример:

$\int_{0}^{2\pi} \dfrac{dx}{(2+\cos x)(3+\cos x)}$

### Решение:

$I=\int_{0}^{2\pi}\dfrac{dx}{(2+\cos x)(3+\cos x)}$

$\dfrac{1}{(2+\cos x)(3+\cos x)}=\dfrac{1}{2+\cos x}-\dfrac{1}{3+\cos x}$
$I=I_1-I_2$
$I_1=\int_{0}^{2\pi}\dfrac{dx}{2+\cos x}$
$I_2=\int_{0}^{2\pi}\dfrac{dx}{3+\cos x}$

$\cos(\pi+t)=-\cos t$
$\int_{0}^{2\pi}\dfrac{dx}{a+\cos x}=2\int_{0}^{\pi}\dfrac{dx}{a+\cos x}\ (a>1)$

$I_1=2\int_{0}^{\pi}\dfrac{dx}{2+\cos x}$
$t=\tan\dfrac{x}{2}$
$\cos x=\dfrac{1-t^2}{1+t^2}$
$dx=\dfrac{2dt}{1+t^2}$
$x:0\to\pi;\ t:0\to+\infty$

$I_1=2\int_{0}^{+\infty}\dfrac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=2\int_{0}^{+\infty}\dfrac{2dt}{3+t^2}=4\int_{0}^{+\infty}\dfrac{dt}{t^2+3}$
$\int\dfrac{dt}{t^2+a^2}=\dfrac{1}{a}\arctan\dfrac{t}{a}$
$I_1=4\cdot\dfrac{1}{\sqrt3}\arctan\dfrac{t}{\sqrt3}\Big|_{0}^{+\infty}=4\cdot\dfrac{1}{\sqrt3}(\dfrac{\pi}{2}-0)=\dfrac{2\pi}{\sqrt3}$

$I_2=2\int_{0}^{\pi}\dfrac{dx}{3+\cos x}$
$t=\tan\dfrac{x}{2}$
$\cos x=\dfrac{1-t^2}{1+t^2}$
$dx=\dfrac{2dt}{1+t^2}$
$x:0\to\pi;\ t:0\to+\infty$

$I_2=2\int_{0}^{+\infty}\dfrac{\frac{2dt}{1+t^2}}{3+\frac{1-t^2}{1+t^2}}=2\int_{0}^{+\infty}\dfrac{2dt}{4+2t^2}=2\int_{0}^{+\infty}\dfrac{dt}{t^2+2}$
$I_2=2\cdot\dfrac{1}{\sqrt2}\arctan\dfrac{t}{\sqrt2}\Big|_{0}^{+\infty}=2\cdot\dfrac{1}{\sqrt2}(\dfrac{\pi}{2}-0)=\dfrac{\pi}{\sqrt2}$

$I=\dfrac{2\pi}{\sqrt3}-\dfrac{\pi}{\sqrt2}$

## 29 номер – Д 2278

### Пример:

$\int_{0}^{\pi} (x\sin x)^{2} \, dx$

### Решение:

$I=\int_{0}^{\pi}(x\sin x)^2dx=\int_{0}^{\pi}x^{2}\sin^{2}x\,dx$

$\sin^{2}x=\dfrac{1-\cos2x}{2}$
$I=\dfrac12\int_{0}^{\pi}x^{2}dx-\dfrac12\int_{0}^{\pi}x^{2}\cos2x\,dx$
$I=\dfrac12.\dfrac{x^{3}}{3}|_{0}^{\pi}-\dfrac12J=\dfrac{\pi^{3}}{6}-\dfrac12J$

$J=\int_{0}^{\pi}x^{2}\cos2x\,dx$
$J=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x^{2};\ du=2x\,dx$
$dv=\cos2x\,dx;\ v=\dfrac12\sin2x$

$J=.\dfrac{x^{2}}{2}\sin2x|_{0}^{\pi}-\int_{0}^{\pi}x\sin2x\,dx=-K$

$K=\int_{0}^{\pi}x\sin2x\,dx$
$K=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x;\ du=dx$
$dv=\sin2x\,dx;\ v=-\dfrac12\cos2x$

$K=.-\dfrac{x}{2}\cos2x|_{0}^{\pi}+\dfrac12\int_{0}^{\pi}\cos2x\,dx$
$\int\cos2x\,dx=\dfrac12\sin2x$
$K=-\dfrac{\pi}{2}\cdot1+\dfrac12.\dfrac12\sin2x|_{0}^{\pi}=-\dfrac{\pi}{2}$

$J=-K=\dfrac{\pi}{2}$
$I=\dfrac{\pi^{3}}{6}-\dfrac12\cdot\dfrac{\pi}{2}=\dfrac{\pi^{3}}{6}-\dfrac{\pi}{4}$

## 40 номер – Д 2395

### Пример:

$v.p. \int_{-\infty}^{+\infty} \text{arccot}x \, dx$

### Решение:

$I=\text{v.p.}\int_{-\infty}^{+\infty}\text{arccot}\,x\,dx=\lim_{A\to+\infty}\int_{-A}^{A}\text{arccot}\,x\,dx$

$\text{arccot}\,x=\dfrac{\pi}{2}-\arctan x$
$I(A)=\int_{-A}^{A}(\dfrac{\pi}{2}-\arctan x)dx=\dfrac{\pi}{2}\int_{-A}^{A}dx-\int_{-A}^{A}\arctan x\,dx$

$\arctan x\ \text{нечётная}$
$\int_{-A}^{A}\arctan x\,dx=0$
$I(A)=\dfrac{\pi}{2}\cdot2A=\pi A$

$I=\lim_{A\to+\infty}\pi A=+\infty$
$\text{v.p. интеграл расходится увы}$

## 42 номер – Д 2398

### Пример:

Площадь
$y=x^{2};x+y=2$

### Решение:

$y=x^2;\ y=2-x$

$x^2=2-x$
$x^2+x-2=0$
$(x+2)(x-1)=0$
$x_1=-2;\ x_2=1$

$S=\int_{-2}^{1}\big((2-x)-x^2\big)\,dx=\int_{-2}^{1}(2-x-x^2)\,dx$
$S=(2x-\dfrac{x^2}{2}-\dfrac{x^3}{3})\Big|_{-2}^{1}$
$S=(2-\dfrac12-\dfrac13)-(-4-\dfrac{4}{2}+\dfrac{8}{3})=\dfrac{7}{6}+\dfrac{10}{3}=\dfrac{9}{2}$

## 44 номер – Д 2400

### Пример:

Площадь
$y=|lg x|; y=0;x=0,1;x=10;$

### Решение:

$S=\int_{0,1}^{10}|lg x|\,dx$
$lg x<0\ (0<x<1);\ lg x>0\ (x>1)$
$S=\int_{0,1}^{1}(-lg x)\,dx+\int_{1}^{10}lg x\,dx$

$lg x=\dfrac{\ln x}{\ln 10}$
$\int lg x\,dx=\dfrac{1}{\ln 10}\int \ln x\,dx=\dfrac{1}{\ln 10}(x\ln x-x)$

$S_2=\int_{1}^{10}lg x\,dx=\dfrac{1}{\ln 10}(x\ln x-x)\Big|_{1}^{10}=10-\dfrac{9}{\ln 10}$
$S_1=\int_{0,1}^{1}(-lg x)\,dx=-\dfrac{1}{\ln 10}(x\ln x-x)\Big|_{0,1}^{1}=-(-\dfrac{1}{\ln 10}+0,1+\dfrac{0,1}{\ln 10})=-0,1+\dfrac{0,9}{\ln 10}$

$S=S_1+S_2=9,9-\dfrac{8,1}{\ln 10}=\dfrac{99}{10}-\dfrac{81}{10\ln 10}$

## 46 номер – Д 2414

### Пример:

Площадь
$x=2t-t^{2};y=2t^{2}-t^{3};$

### Решение:

$x=t(2-t);\ y=t^{2}(2-t)$
$t=0;\ x=0;\ y=0$
$t=2;\ x=0;\ y=0$

$S=\dfrac12\int\limits_{0}^{2}(x\dfrac{dy}{dt}-y\dfrac{dx}{dt})dt$

$y=tx$
$\dfrac{dy}{dt}=x+t\dfrac{dx}{dt}$
$x\dfrac{dy}{dt}-y\dfrac{dx}{dt}=x(x+t\dfrac{dx}{dt})-tx\dfrac{dx}{dt}=x^{2}$

$S=\dfrac12\int\limits_{0}^{2}x^{2}dt=\dfrac12\int\limits_{0}^{2}(t(2-t))^{2}dt=\dfrac12\int\limits_{0}^{2}(4t^{2}-4t^{3}+t^{4})dt$
$S=\dfrac12(\dfrac{4}{3}t^{3}-t^{4}+\dfrac{1}{5}t^{5})\Big|_{0}^{2}=\dfrac12(\dfrac{32}{3}-16+\dfrac{32}{5})=\dfrac12\cdot\dfrac{16}{15}=\dfrac{8}{15}$

## 48 номер – Д 2418

### Пример:

Площадь
$r^{2}=a^{2}\cos 2\phi \, (лемниската)$

### Решение:

$r^{2}=a^{2}\cos2\phi$
$\cos2\phi\ge0;\ \phi\in[-\dfrac{\pi}{4};\dfrac{\pi}{4}] \ \text{(одна петля)}$

$S_1=\dfrac12\int\limits_{-\pi/4}^{\pi/4}r^{2}d\phi=\dfrac12\int\limits_{-\pi/4}^{\pi/4}a^{2}\cos2\phi\,d\phi$
$S_1=\dfrac{a^{2}}{2}\cdot\dfrac12\sin2\phi\Big|_{-\pi/4}^{\pi/4}=\dfrac{a^{2}}{4}(\sin\dfrac{\pi}{2}-\sin(-\dfrac{\pi}{2}))=\dfrac{a^{2}}{4}(1-(-1))=\dfrac{a^{2}}{2}$

$S=2S_1=a^{2}$

## 50 номер – Д 2431

### Пример:

Длины дуг кривой
$y=x^{\frac{3}{2}}; (0 \leq x \leq 4)$

### Решение:

$l=\int\limits_{0}^{4}\sqrt{1+(y')^{2}}\,dx$
$y=x^{\frac32};\ y'=\dfrac32x^{\frac12}$
$(y')^{2}=\dfrac{9}{4}x$
$l=\int\limits_{0}^{4}\sqrt{1+\dfrac{9}{4}x}\,dx=\dfrac12\int\limits_{0}^{4}\sqrt{9x+4}\,dx$

$u=9x+4;\ du=9dx;\ dx=\dfrac{du}{9}$
$x=0;\ u=4$
$x=4;\ u=40$

$l=\dfrac12\int\limits_{4}^{40}\sqrt{u}\cdot\dfrac{du}{9}=\dfrac{1}{18}\int\limits_{4}^{40}u^{\frac12}du=\dfrac{1}{18}\cdot\dfrac{2}{3}u^{\frac32}\Big|_{4}^{40}=\dfrac{1}{27}(40^{\frac32}-4^{\frac32})$
$40^{\frac32}=40\sqrt{40}=80\sqrt{10};\ 4^{\frac32}=8$
$l=\dfrac{1}{27}(80\sqrt{10}-8)=\dfrac{8}{27}(10\sqrt{10}-1)$

## 52 номер – Д 2433

### Пример:

Длины дуг кривой
$y=a\cosh \dfrac{x}{a}; \text{от точки A(0,a) до точки B(b,h)}$

### Решение:

$a>0$
$l=\int\limits_{0}^{b}\sqrt{1+(y')^{2}}\,dx$
$y=a\cosh\dfrac{x}{a}$
$y'=a(\cosh\dfrac{x}{a})'=a\cdot\sinh\dfrac{x}{a}\cdot\dfrac{1}{a}=\sinh\dfrac{x}{a}$

$1+(y')^{2}=1+\sinh^{2}\dfrac{x}{a}=\cosh^{2}\dfrac{x}{a}$
$\sqrt{1+(y')^{2}}=\cosh\dfrac{x}{a}$

$l=\int\limits_{0}^{b}\cosh\dfrac{x}{a}\,dx=a\sinh\dfrac{x}{a}\Big|_{0}^{b}=a\sinh\dfrac{b}{a}$

$h=y(b)=a\cosh\dfrac{b}{a}$
$\sinh\dfrac{b}{a}=\sqrt{\cosh^{2}\dfrac{b}{a}-1}=\sqrt{(\dfrac{h}{a})^{2}-1}=\dfrac{\sqrt{h^{2}-a^{2}}}{a}$
$l=a\sinh\dfrac{b}{a}=\sqrt{h^{2}-a^{2}}$

## 54 номер – Д 2462

### Пример:

Объём
$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1;z=\dfrac{c}{a}x;z=0;$

### Решение:

$V=\iiint\limits_{(V)}dV=\iint\limits_{D}(z_{\text{верх}}-z_{\text{низ}})\,dS$

$D:\ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}\le1$
$z_{\text{низ}}=0;\ z_{\text{верх}}=\dfrac{c}{a}x$

$z_{\text{верх}}\ge z_{\text{низ}};\ \dfrac{c}{a}x\ge0;\ x\ge0$
$D_{1}=D\cap\{x\ge0\}$

$V=\iint\limits_{D_{1}}\dfrac{c}{a}x\,dS=\dfrac{c}{a}\iint\limits_{D_{1}}x\,dS$

$x=a r\cos t;\ y=b r\sin t;\ 0\le r\le1;\ -\dfrac{\pi}{2}\le t\le\dfrac{\pi}{2}$
$dS=ab\,r\,dr\,dt$

$\iint\limits_{D_{1}}x\,dS=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{0}^{1}(a r\cos t)\,ab\,r\,dr\,dt=a^{2}b\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t\,dt\int\limits_{0}^{1}r^{2}\,dr$
$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t\,dt=.\sin t|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=2;\ \int\limits_{0}^{1}r^{2}\,dr=.\dfrac{r^{3}}{3}|_{0}^{1}=\dfrac13$
$\iint\limits_{D_{1}}x\,dS=a^{2}b\cdot2\cdot\dfrac13=\dfrac{2a^{2}b}{3}$

$V=\dfrac{c}{a}\cdot\dfrac{2a^{2}b}{3}=\dfrac{2abc}{3}$

## 56 номер – Д 2464

### Пример:

Объём
$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}-\dfrac{z^{2}}{c^{2}}=1;z=\pm c$

### Решение:

$V=\int\limits_{-c}^{c}S(z)\,dz$

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}-\dfrac{z^{2}}{c^{2}}=1$
$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1+\dfrac{z^{2}}{c^{2}}$

$S(z)=\pi\cdot a\sqrt{1+\dfrac{z^{2}}{c^{2}}}\cdot b\sqrt{1+\dfrac{z^{2}}{c^{2}}}=\pi ab(1+\dfrac{z^{2}}{c^{2}})$

$V=\int\limits_{-c}^{c}\pi ab(1+\dfrac{z^{2}}{c^{2}})dz=\pi ab(\int\limits_{-c}^{c}dz+\dfrac{1}{c^{2}}\int\limits_{-c}^{c}z^{2}dz)$
$\int\limits_{-c}^{c}dz=2c;\ \int\limits_{-c}^{c}z^{2}dz=.\dfrac{z^{3}}{3}|_{-c}^{c}=\dfrac{2c^{3}}{3}$
$V=\pi ab(2c+\dfrac{1}{c^{2}}\cdot\dfrac{2c^{3}}{3})=\pi ab(2c+\dfrac{2c}{3})=\dfrac{8\pi abc}{3}$

## 58 номер – Д 2666
АЦЦЦККИИИИИЙ НОМЕР он ещё и последний
и именно поэтому я его делать НЕ БУДУ  :D