Całki podwójne
- interaktywna sesja notebooka
import ipywidgets as widgets
from IPython.display import display, Markdown, Latex
import generator_zadan.generatory as gz
print(gz.__version__)
0.2.11
ile_zadan_przykladowych = 10
Po trójkącie
zadanie = gz.calka_podwojna(typ=1, nr_zadania=1)
zadanie
! Tworzę katalog pics
('Obliczyć $$\\iint\\limits_D\\left(\\frac{3 y}{2} + 1\\right)dx\\,dy$$ gdzie $D$ - trójkąt $ABC$ dla $A=(-2,3), B=(0,0), C=(0,-3)$',
'\\includegraphics[width=60mm]{../pics/calka_podwojna1x}\n\t\\includegraphics[width=60mm]{../pics/calka_podwojna1y}\\\\\n\tWzględem $Ox:$ $\\int\\limits_{-2}^{0}\\left(\\int\\limits_{- 3 x - 3}^{- \\frac{3 x}{2}}\\left(\\frac{3 y}{2} + 1\\right)dy\\right)dx\n\t=\\ldots=\\int\\limits_{-2}^{0}\\left(- \\frac{81 x^{2}}{16} - 12 x - \\frac{15}{4}\\right)dx = \\ldots = 3$ \\\\\n\tWzględem $Oy:$ $\\int\\limits_{-3}^{0}\\left(\\int\\limits_{- \\frac{y}{3} - 1}^{0}\\left(\\frac{3 y}{2} + 1\\right)dx\\right)dy + \\int\\limits_{0}^{3}\\left(\\int\\limits_{- \\frac{y}{3} - 1}^{- \\frac{2 y}{3}}\\left(\\frac{3 y}{2} + 1\\right)dx\\right)dy\n\t=\\ldots=\\int\\limits_{-3}^{0}\\left( \\frac{y^{2}}{2} + \\frac{11 y}{6} + 1 \\right)dy + \\int\\limits_{0}^{3}\\left( - \\frac{y^{2}}{2} + \\frac{7 y}{6} + 1 \\right)dy = \\ldots = 3$ \\\\')
Show code cell source
print("\033[34m** Zadanie **" + '*' * 81 + '\033[0m')
print(zadanie[0])
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
print(zadanie[1])
print("\033[32m*\033[0m" * 95)
** Zadanie ***********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{3 y}{2} + 1\right)dx\,dy$$ gdzie $D$ - trójkąt $ABC$ dla $A=(-2,3), B=(0,0), C=(0,-3)$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
\includegraphics[width=60mm]{../pics/calka_podwojna1x}
\includegraphics[width=60mm]{../pics/calka_podwojna1y}\\
Względem $Ox:$ $\int\limits_{-2}^{0}\left(\int\limits_{- 3 x - 3}^{- \frac{3 x}{2}}\left(\frac{3 y}{2} + 1\right)dy\right)dx
=\ldots=\int\limits_{-2}^{0}\left(- \frac{81 x^{2}}{16} - 12 x - \frac{15}{4}\right)dx = \ldots = 3$ \\
Względem $Oy:$ $\int\limits_{-3}^{0}\left(\int\limits_{- \frac{y}{3} - 1}^{0}\left(\frac{3 y}{2} + 1\right)dx\right)dy + \int\limits_{0}^{3}\left(\int\limits_{- \frac{y}{3} - 1}^{- \frac{2 y}{3}}\left(\frac{3 y}{2} + 1\right)dx\right)dy
=\ldots=\int\limits_{-3}^{0}\left( \frac{y^{2}}{2} + \frac{11 y}{6} + 1 \right)dy + \int\limits_{0}^{3}\left( - \frac{y^{2}}{2} + \frac{7 y}{6} + 1 \right)dy = \ldots = 3$ \\
***********************************************************************************************
Show code cell source
for i in range(1, ile_zadan_przykladowych + 1):
zadanie = gz.calka_podwojna(typ=1, nr_zadania=i)
print(f"\033[34m** Zadanie {i} **" + '*' * 80 + '\033[0m')
display(Markdown(zadanie[0].replace('\\,', '').replace('$', '*', ).replace('**', '$$', 2)))
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
img1 = open(f'pics/calka_podwojna{i}x.png', 'rb').read()
wi1 = widgets.Image(value=img1, format='jpg', width=300, height=400)
img2 = open(f'pics/calka_podwojna{i}y.png', 'rb').read()
wi2 = widgets.Image(value=img2, format='jpg', width=300, height=400)
a = [wi1, wi2]
wid = widgets.HBox(a)
display(wid)
if len(zadanie[1].split('$')) == 5:
display(Markdown(f'Względem *{zadanie[1].split("$")[1]}*'))
display(Markdown('$$' + zadanie[1].split('$')[3].split('$')[0] + '$$'))
else:
display(Markdown(f'Względem *{zadanie[1].split("$")[1]}*'))
display(Markdown('$$' + zadanie[1].split('$')[3].split('$')[0] + '$$'))
display(Markdown(f'Względem *{zadanie[1].split("$")[5]}*'))
display(Markdown('$$' + zadanie[1].split('$')[7].split('$')[0] + '$$'))
print("\033[32m*\033[0m" * 95)
** Zadanie 1 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 x + \frac{3 y}{2} + 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,1), B=(-2,-2), C=(-2,1)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-2}^{0}\left(\int\limits_{\frac{3 x}{2} + 1}^{1}\left(2 x + \frac{3 y}{2} + 1\right)dy\right)dx =\ldots=\int\limits_{-2}^{0}\left(- \frac{75 x^{2}}{16} - \frac{15 x}{4}\right)dx = \ldots = -5$$
Względem Oy:
$$\int\limits_{-2}^{1}\left(\int\limits_{-2}^{\frac{2 y}{3} - \frac{2}{3}}\left(2 x + \frac{3 y}{2} + 1\right)dx\right)dy =\ldots=\int\limits_{-2}^{1}\left(\frac{13 y^{2}}{9} + \frac{16 y}{9} - \frac{20}{9}\right)dy = \ldots = -5$$
***********************************************************************************************
** Zadanie 2 **********************************************************************************
Obliczyć $$\iint\limits_D\left(x + 2 y\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,2), B=(0,0), C=(4,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{4}\left(\int\limits_{0}^{2 - \frac{x}{2}}\left(x + 2 y\right)dy\right)dx =\ldots=\int\limits_{0}^{4}\left(4 - \frac{x^{2}}{4}\right)dx = \ldots = \frac{32}{3}$$
Względem Oy:
$$\int\limits_{0}^{2}\left(\int\limits_{0}^{4 - 2 y}\left(x + 2 y\right)dx\right)dy =\ldots=\int\limits_{0}^{2}\left(8 - 2 y^{2}\right)dy = \ldots = \frac{32}{3}$$
***********************************************************************************************
** Zadanie 3 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 y\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-3,-2), B=(2,0), C=(0,-2)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-3}^{0}\left(\int\limits_{-2}^{\frac{2 x}{5} - \frac{4}{5}}\left(- 2 y\right)dy\right)dx + \int\limits_{0}^{2}\left(\int\limits_{x - 2}^{\frac{2 x}{5} - \frac{4}{5}}\left(- 2 y\right)dy\right)dx =\ldots=\int\limits_{-3}^{0}\left( - \frac{4 x^{2}}{25} + \frac{16 x}{25} + \frac{84}{25} \right)dx + \int\limits_{0}^{2}\left( \frac{21 x^{2}}{25} - \frac{84 x}{25} + \frac{84}{25} \right)dx = \ldots = 8$$
Względem Oy:
$$\int\limits_{-2}^{0}\left(\int\limits_{\frac{5 y}{2} + 2}^{y + 2}\left(- 2 y\right)dx\right)dy =\ldots=\int\limits_{-2}^{0}\left(3 y^{2}\right)dy = \ldots = 8$$
***********************************************************************************************
** Zadanie 4 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 - 3 y\right)dxdy$$ gdzie D - trójkąt ABC dla A=(1,0), B=(2,2), C=(0,-1)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{1}\left(\int\limits_{x - 1}^{\frac{3 x}{2} - 1}\left(2 - 3 y\right)dy\right)dx + \int\limits_{1}^{2}\left(\int\limits_{2 x - 2}^{\frac{3 x}{2} - 1}\left(2 - 3 y\right)dy\right)dx =\ldots=\int\limits_{0}^{1}\left( - \frac{15 x^{2}}{8} + \frac{5 x}{2} \right)dx + \int\limits_{1}^{2}\left( \frac{21 x^{2}}{8} - \frac{17 x}{2} + \frac{13}{2} \right)dx = \ldots = \frac{1}{2}$$
Względem Oy:
$$\int\limits_{-1}^{0}\left(\int\limits_{\frac{2 y}{3} + \frac{2}{3}}^{y + 1}\left(2 - 3 y\right)dx\right)dy + \int\limits_{0}^{2}\left(\int\limits_{\frac{2 y}{3} + \frac{2}{3}}^{\frac{y}{2} + 1}\left(2 - 3 y\right)dx\right)dy =\ldots=\int\limits_{-1}^{0}\left( - y^{2} - \frac{y}{3} + \frac{2}{3} \right)dy + \int\limits_{0}^{2}\left( \frac{y^{2}}{2} - \frac{4 y}{3} + \frac{2}{3} \right)dy = \ldots = \frac{1}{2}$$
***********************************************************************************************
** Zadanie 5 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 x - 3 y - 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,0), B=(0,2), C=(4,-3)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{4}\left(\int\limits_{- \frac{3 x}{4}}^{2 - \frac{5 x}{4}}\left(2 x - 3 y - 1\right)dy\right)dx =\ldots=\int\limits_{0}^{4}\left(- \frac{5 x^{2}}{2} + 12 x - 8\right)dx = \ldots = \frac{32}{3}$$
Względem Oy:
$$\int\limits_{-3}^{0}\left(\int\limits_{- \frac{4 y}{3}}^{\frac{8}{5} - \frac{4 y}{5}}\left(2 x - 3 y - 1\right)dx\right)dy + \int\limits_{0}^{2}\left(\int\limits_{0}^{\frac{8}{5} - \frac{4 y}{5}}\left(2 x - 3 y - 1\right)dx\right)dy =\ldots=\int\limits_{-3}^{0}\left( - \frac{616 y^{2}}{225} - \frac{592 y}{75} + \frac{24}{25} \right)dy + \int\limits_{0}^{2}\left( \frac{76 y^{2}}{25} - \frac{164 y}{25} + \frac{24}{25} \right)dy = \ldots = \frac{32}{3}$$
***********************************************************************************************
** Zadanie 6 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x + \frac{y}{2} - 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-1,4), B=(1,0), C=(0,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{0}\left(\int\limits_{- 4 x}^{2 - 2 x}\left(- x + \frac{y}{2} - 1\right)dy\right)dx + \int\limits_{0}^{1}\left(\int\limits_{0}^{2 - 2 x}\left(- x + \frac{y}{2} - 1\right)dy\right)dx =\ldots=\int\limits_{-1}^{0}\left( - 5 x^{2} - 6 x - 1 \right)dx + \int\limits_{0}^{1}\left( 3 x^{2} - 2 x - 1 \right)dx = \ldots = - \frac{2}{3}$$
Względem Oy:
$$\int\limits_{0}^{4}\left(\int\limits_{- \frac{y}{4}}^{1 - \frac{y}{2}}\left(- x + \frac{y}{2} - 1\right)dx\right)dy =\ldots=\int\limits_{0}^{4}\left(- \frac{7 y^{2}}{32} + \frac{5 y}{4} - \frac{3}{2}\right)dy = \ldots = - \frac{2}{3}$$
***********************************************************************************************
** Zadanie 7 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{x}{2} + 3 y - 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(4,0), B=(-1,1), C=(0,-2)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{0}\left(\int\limits_{- 3 x - 2}^{\frac{4}{5} - \frac{x}{5}}\left(\frac{x}{2} + 3 y - 1\right)dy\right)dx + \int\limits_{0}^{4}\left(\int\limits_{\frac{x}{2} - 2}^{\frac{4}{5} - \frac{x}{5}}\left(\frac{x}{2} + 3 y - 1\right)dy\right)dx =\ldots=\int\limits_{-1}^{0}\left( - \frac{301 x^{2}}{25} - \frac{497 x}{25} - \frac{196}{25} \right)dx + \int\limits_{0}^{4}\left( - \frac{133 x^{2}}{200} + \frac{231 x}{50} - \frac{196}{25} \right)dx = \ldots = - \frac{21}{2}$$
Względem Oy:
$$\int\limits_{-2}^{0}\left(\int\limits_{- \frac{y}{3} - \frac{2}{3}}^{2 y + 4}\left(\frac{x}{2} + 3 y - 1\right)dx\right)dy + \int\limits_{0}^{1}\left(\int\limits_{- \frac{y}{3} - \frac{2}{3}}^{4 - 5 y}\left(\frac{x}{2} + 3 y - 1\right)dx\right)dy =\ldots=\int\limits_{-2}^{0}\left( \frac{287 y^{2}}{36} + \frac{140 y}{9} - \frac{7}{9} \right)dy + \int\limits_{0}^{1}\left( - \frac{70 y^{2}}{9} + \frac{77 y}{9} - \frac{7}{9} \right)dy = \ldots = - \frac{21}{2}$$
***********************************************************************************************
** Zadanie 8 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{x}{2}\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,4), B=(0,0), C=(1,-2)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{1}\left(\int\limits_{- 2 x}^{4 - 6 x}\left(\frac{x}{2}\right)dy\right)dx =\ldots=\int\limits_{0}^{1}\left(- 2 x^{2} + 2 x\right)dx = \ldots = \frac{1}{3}$$
Względem Oy:
$$\int\limits_{-2}^{0}\left(\int\limits_{- \frac{y}{2}}^{\frac{2}{3} - \frac{y}{6}}\left(\frac{x}{2}\right)dx\right)dy + \int\limits_{0}^{4}\left(\int\limits_{0}^{\frac{2}{3} - \frac{y}{6}}\left(\frac{x}{2}\right)dx\right)dy =\ldots=\int\limits_{-2}^{0}\left( - \frac{y^{2}}{18} - \frac{y}{18} + \frac{1}{9} \right)dy + \int\limits_{0}^{4}\left( \frac{y^{2}}{144} - \frac{y}{18} + \frac{1}{9} \right)dy = \ldots = \frac{1}{3}$$
***********************************************************************************************
** Zadanie 9 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x - y + 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(1,0), B=(0,0), C=(0,1)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{1}\left(\int\limits_{0}^{1 - x}\left(- x - y + 1\right)dy\right)dx =\ldots=\int\limits_{0}^{1}\left(\frac{x^{2}}{2} - x + \frac{1}{2}\right)dx = \ldots = \frac{1}{6}$$
Względem Oy:
$$\int\limits_{0}^{1}\left(\int\limits_{0}^{1 - y}\left(- x - y + 1\right)dx\right)dy =\ldots=\int\limits_{0}^{1}\left(\frac{y^{2}}{2} - y + \frac{1}{2}\right)dy = \ldots = \frac{1}{6}$$
***********************************************************************************************
** Zadanie 10 **********************************************************************************
Obliczyć $$\iint\limits_D\left(3 x + 3 y + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-3,3), B=(-1,3), C=(0,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-3}^{-1}\left(\int\limits_{- x}^{3}\left(3 x + 3 y + 2\right)dy\right)dx + \int\limits_{-1}^{0}\left(\int\limits_{- x}^{- 3 x}\left(3 x + 3 y + 2\right)dy\right)dx =\ldots=\int\limits_{-3}^{-1}\left( \frac{3 x^{2}}{2} + 11 x + \frac{39}{2} \right)dx + \int\limits_{-1}^{0}\left( 6 x^{2} - 4 x \right)dx = \ldots = 12$$
Względem Oy:
$$\int\limits_{0}^{3}\left(\int\limits_{- y}^{- \frac{y}{3}}\left(3 x + 3 y + 2\right)dx\right)dy =\ldots=\int\limits_{0}^{3}\left(\frac{2 y^{2}}{3} + \frac{4 y}{3}\right)dy = \ldots = 12$$
***********************************************************************************************
Po obszarze ograniczonym krzywymi
zadanie = gz.calka_podwojna(typ=2, nr_zadania=1)
zadanie
('Obliczyć $$\\iint\\limits_D\\left(- 2 x + 3 y\\right)dx\\,dy$$ gdzie $D$ - obszar ograniczony krzywymi $y=\\frac{x^{2}}{2} + 1$, oraz $y=- 3 x - 3$',
'\\includegraphics[width=45mm]{../pics/calka_podwojna1x}\n\t\\includegraphics[width=45mm]{../pics/calka_podwojna1y}\\\\\n\tWzględem $Ox:$ $\\int\\limits_{-4}^{-2}\\left(\\int\\limits_{\\frac{x^{2}}{2} + 1}^{- 3 x - 3}\\left(- 2 x + 3 y\\right)dy\\right)dx\n\t=\\ldots=\\int\\limits_{-4}^{-2}\\left(- \\frac{3 x^{4}}{8} + x^{3} + 18 x^{2} + 35 x + 12\\right)dx = \\ldots = \\frac{78}{5}$ \\\\')
Show code cell source
print("\033[34m** Zadanie **" + '*' * 81 + '\033[0m')
print(zadanie[0])
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
print(zadanie[1])
print("\033[32m*\033[0m" * 95)
** Zadanie ***********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + 3 y\right)dx\,dy$$ gdzie $D$ - obszar ograniczony krzywymi $y=\frac{x^{2}}{2} + 1$, oraz $y=- 3 x - 3$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
\includegraphics[width=45mm]{../pics/calka_podwojna1x}
\includegraphics[width=45mm]{../pics/calka_podwojna1y}\\
Względem $Ox:$ $\int\limits_{-4}^{-2}\left(\int\limits_{\frac{x^{2}}{2} + 1}^{- 3 x - 3}\left(- 2 x + 3 y\right)dy\right)dx
=\ldots=\int\limits_{-4}^{-2}\left(- \frac{3 x^{4}}{8} + x^{3} + 18 x^{2} + 35 x + 12\right)dx = \ldots = \frac{78}{5}$ \\
***********************************************************************************************
Show code cell source
for i in range(ile_zadan_przykladowych, 2 * ile_zadan_przykladowych + 1):
zadanie = gz.calka_podwojna(typ=2, nr_zadania=i)
print(f"\033[34m** Zadanie {i} **" + '*' * 80 + '\033[0m')
display(Markdown(
zadanie[0].split('krzywymi')[0].replace('\\,', '').replace('$', '*', ).replace('**', '$$', 2) + 'krzywymi'))
display(Markdown('$$' + zadanie[0].split('$')[7] + '\\quad \\text{' + zadanie[0].split('$')[8][1:] + '} \\quad ' +
zadanie[0].split('$')[9] + '$$'))
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
img1 = open(f'pics/calka_podwojna{i}x.png', 'rb').read()
wi1 = widgets.Image(value=img1, format='jpg', width=300, height=400)
img2 = open(f'pics/calka_podwojna{i}y.png', 'rb').read()
wi2 = widgets.Image(value=img2, format='jpg', width=300, height=400)
a = [wi1, wi2]
wid = widgets.HBox(a)
display(wid)
if len(zadanie[1].split('$')) == 5:
display(Markdown(f'Względem *{zadanie[1].split("$")[1]}*'))
display(Markdown('$$' + zadanie[1].split('$')[3].split('$')[0] + '$$'))
else:
display(Markdown(f'Względem *{zadanie[1].split("$")[1]}*'))
display(Markdown('$$' + zadanie[1].split('$')[3].split('$')[0] + '$$'))
display(Markdown(f'Względem *{zadanie[1].split("$")[5]}*'))
display(Markdown('$$' + zadanie[1].split('$')[7].split('$')[0] + '$$'))
print("\033[32m*\033[0m" * 95)
** Zadanie 10 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=- y^{2}\quad \text{ oraz } \quad x=- 2 y$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{0}^{2}\left(\int\limits_{- 2 y}^{- y^{2}}\left(- y\right)dx\right)dy =\ldots=\int\limits_{0}^{2}\left(y^{3} - 2 y^{2}\right)dy = \ldots = - \frac{4}{3}$$
***********************************************************************************************
** Zadanie 11 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=\frac{x^{2}}{2} + 3 x - 2\quad \text{ oraz } \quad y=3 x$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-2}^{2}\left(\int\limits_{\frac{x^{2}}{2} + 3 x - 2}^{3 x}\left(- 2 x + y\right)dy\right)dx =\ldots=\int\limits_{-2}^{2}\left(- \frac{x^{4}}{8} - \frac{x^{3}}{2} + x^{2} + 2 x - 2\right)dx = \ldots = - \frac{64}{15}$$
Względem Oy:
$$\int\limits_{-6}^{6}\left(\int\limits_{\frac{y}{3}}^{\sqrt{2 y + 13} - 3}\left(- 2 x + y\right)dx\right)dy =\ldots=\int\limits_{-6}^{6}\left(- \frac{2 y^{2}}{9} + y \sqrt{2 y + 13} - 5 y + 6 \sqrt{2 y + 13} - 22\right)dy = \ldots = - \frac{64}{15}$$
***********************************************************************************************
** Zadanie 12 **********************************************************************************
Obliczyć $$\iint\limits_D\left(1 - x\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=- x^{2} + 3 x + 1\quad \text{ oraz } \quad y=2 x - 1$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{2}\left(\int\limits_{2 x - 1}^{- x^{2} + 3 x + 1}\left(1 - x\right)dy\right)dx =\ldots=\int\limits_{-1}^{2}\left(x^{3} - 2 x^{2} - x + 2\right)dx = \ldots = \frac{9}{4}$$
***********************************************************************************************
** Zadanie 13 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 x + 2 y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=- x^{2} - 3 x + 1\quad \text{ oraz } \quad y=- 3 x$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{1}\left(\int\limits_{- 3 x}^{- x^{2} - 3 x + 1}\left(2 x + 2 y\right)dy\right)dx =\ldots=\int\limits_{-1}^{1}\left(x^{4} + 4 x^{3} - 2 x^{2} - 4 x + 1\right)dx = \ldots = \frac{16}{15}$$
***********************************************************************************************
** Zadanie 14 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x + 2 y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=\frac{3 y^{2}}{2} - 1\quad \text{ oraz } \quad x=3 y - 1$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{0}^{2}\left(\int\limits_{\frac{3 y^{2}}{2} - 1}^{3 y - 1}\left(- x + 2 y\right)dx\right)dy =\ldots=\int\limits_{0}^{2}\left(\frac{9 y^{4}}{8} - 3 y^{3} + 3 y\right)dy = \ldots = \frac{6}{5}$$
***********************************************************************************************
** Zadanie 15 **********************************************************************************
Obliczyć $$\iint\limits_D\left(3 y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=\frac{x^{2}}{2}\quad \text{ oraz } \quad y=\frac{x}{2} + 1$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{2}\left(\int\limits_{\frac{x^{2}}{2}}^{\frac{x}{2} + 1}\left(3 y\right)dy\right)dx =\ldots=\int\limits_{-1}^{2}\left(- \frac{3 x^{4}}{8} + \frac{3 x^{2}}{8} + \frac{3 x}{2} + \frac{3}{2}\right)dx = \ldots = \frac{27}{5}$$
Względem Oy:
$$\int\limits_{0}^{1/2}\left(\int\limits_{- \sqrt{2} \sqrt{y}}^{\sqrt{2} \sqrt{y}}\left(3 y\right)dx\right)dy + \int\limits_{1/2}^{2 }\left(\int\limits_{2 y - 2}^{\sqrt{2} \sqrt{y}}\left(3 y\right)dx\right)dy =\ldots=\int\limits_{0}^{1/2}\left(6 \sqrt{2} y^{\frac{3}{2}}\right)dy + \int\limits_{1/2}^{2}\left(3 \sqrt{2} y^{\frac{3}{2}} - 6 y^{2} + 6 y\right)dx = \ldots = \frac{27}{5}$$
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** Zadanie 16 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=2 x^{2}\quad \text{ oraz } \quad y=- 2 x$$
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** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{0}\left(\int\limits_{2 x^{2}}^{- 2 x}\left(- 2 x + y\right)dy\right)dx =\ldots=\int\limits_{-1}^{0}\left(- 2 x^{4} + 4 x^{3} + 6 x^{2}\right)dx = \ldots = \frac{3}{5}$$
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** Zadanie 17 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 3 x - 2 y + 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=- x^{2} + 2 x\quad \text{ oraz } \quad y=2 x - 1$$
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** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{1}\left(\int\limits_{2 x - 1}^{- x^{2} + 2 x}\left(- 3 x - 2 y + 1\right)dy\right)dx =\ldots=\int\limits_{-1}^{1}\left(- x^{4} + 7 x^{3} - x^{2} - 7 x + 2\right)dx = \ldots = \frac{44}{15}$$
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** Zadanie 18 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + 3 y + 2\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=- 3 x^{2}\quad \text{ oraz } \quad y=- 3 x$$
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** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{1}\left(\int\limits_{- 3 x}^{- 3 x^{2}}\left(- 2 x + 3 y + 2\right)dy\right)dx =\ldots=\int\limits_{0}^{1}\left(\frac{27 x^{4}}{2} + 6 x^{3} - \frac{51 x^{2}}{2} + 6 x\right)dx = \ldots = - \frac{13}{10}$$
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** Zadanie 19 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 3 y - 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=3 - x^{2}\quad \text{ oraz } \quad y=2 x$$
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** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-3}^{1}\left(\int\limits_{2 x}^{3 - x^{2}}\left(- 3 y - 1\right)dy\right)dx =\ldots=\int\limits_{-3}^{1}\left(- \frac{3 x^{4}}{2} + 16 x^{2} + 2 x - \frac{33}{2}\right)dx = \ldots = \frac{32}{15}$$
***********************************************************************************************
** Zadanie 20 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x + y + 2\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=\frac{x^{2}}{2} + 3 x\quad \text{ oraz } \quad y=- 2 x$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-10}^{0}\left(\int\limits_{\frac{x^{2}}{2} + 3 x}^{- 2 x}\left(- x + y + 2\right)dy\right)dx =\ldots=\int\limits_{-10}^{0}\left(- \frac{x^{4}}{8} - x^{3} + \frac{3 x^{2}}{2} - 10 x\right)dx = \ldots = 1000$$
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