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__)
0.2.10
ile_zadan_przykladowych = 10
Po trójkącie
zadanie = gz.calka_podwojna(typ=1, nr_zadania=1)
zadanie
! Tworzę katalog pics
('Obliczyć $$\\iint\\limits_D\\left(\\frac{3 x}{2} + \\frac{y}{2} + 1\\right)dx\\,dy$$ gdzie $D$ - trójkąt $ABC$ dla $A=(0,2), B=(1,-3), C=(-3,0)$',
'\\includegraphics[width=60mm]{../pics/calka_podwojna1x}\n\t\\includegraphics[width=60mm]{../pics/calka_podwojna1y}\\\\\n\tWzględem $Ox:$ $\\int\\limits_{-3}^{0}\\left(\\int\\limits_{- \\frac{3 x}{4} - \\frac{9}{4}}^{\\frac{2 x}{3} + 2}\\left(\\frac{3 x}{2} + \\frac{y}{2} + 1\\right)dy\\right)dx + \\int\\limits_{0}^{1}\\left(\\int\\limits_{- \\frac{3 x}{4} - \\frac{9}{4}}^{2 - 5 x}\\left(\\frac{3 x}{2} + \\frac{y}{2} + 1\\right)dy\\right)dx\n\t=\\ldots=\\int\\limits_{-3}^{0}\\left( \\frac{1207 x^{2}}{576} + \\frac{731 x}{96} + \\frac{255}{64} \\right)dx + \\int\\limits_{0}^{1}\\left( - \\frac{17 x^{2}}{64} - \\frac{119 x}{32} + \\frac{255}{64} \\right)dx = \\ldots = - \\frac{17}{12}$ \\\\\n\tWzględem $Oy:$ $\\int\\limits_{-3}^{0}\\left(\\int\\limits_{- \\frac{4 y}{3} - 3}^{\\frac{2}{5} - \\frac{y}{5}}\\left(\\frac{3 x}{2} + \\frac{y}{2} + 1\\right)dx\\right)dy + \\int\\limits_{0}^{2}\\left(\\int\\limits_{\\frac{3 y}{2} - 3}^{\\frac{2}{5} - \\frac{y}{5}}\\left(\\frac{3 x}{2} + \\frac{y}{2} + 1\\right)dx\\right)dy\n\t=\\ldots=\\int\\limits_{-3}^{0}\\left( - \\frac{221 y^{2}}{300} - \\frac{493 y}{150} - \\frac{323}{100} \\right)dy + \\int\\limits_{0}^{2}\\left( - \\frac{1003 y^{2}}{400} + \\frac{663 y}{100} - \\frac{323}{100} \\right)dy = \\ldots = - \\frac{17}{12}$ \\\\')
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 $$\iint\limits_D\left(\frac{3 x}{2} + \frac{y}{2} + 1\right)dx\,dy$$ gdzie $D$ - trójkąt $ABC$ dla $A=(0,2), B=(1,-3), C=(-3,0)$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
\includegraphics[width=60mm]{../pics/calka_podwojna1x}
\includegraphics[width=60mm]{../pics/calka_podwojna1y}\\
Względem $Ox:$ $\int\limits_{-3}^{0}\left(\int\limits_{- \frac{3 x}{4} - \frac{9}{4}}^{\frac{2 x}{3} + 2}\left(\frac{3 x}{2} + \frac{y}{2} + 1\right)dy\right)dx + \int\limits_{0}^{1}\left(\int\limits_{- \frac{3 x}{4} - \frac{9}{4}}^{2 - 5 x}\left(\frac{3 x}{2} + \frac{y}{2} + 1\right)dy\right)dx
=\ldots=\int\limits_{-3}^{0}\left( \frac{1207 x^{2}}{576} + \frac{731 x}{96} + \frac{255}{64} \right)dx + \int\limits_{0}^{1}\left( - \frac{17 x^{2}}{64} - \frac{119 x}{32} + \frac{255}{64} \right)dx = \ldots = - \frac{17}{12}$ \\
Względem $Oy:$ $\int\limits_{-3}^{0}\left(\int\limits_{- \frac{4 y}{3} - 3}^{\frac{2}{5} - \frac{y}{5}}\left(\frac{3 x}{2} + \frac{y}{2} + 1\right)dx\right)dy + \int\limits_{0}^{2}\left(\int\limits_{\frac{3 y}{2} - 3}^{\frac{2}{5} - \frac{y}{5}}\left(\frac{3 x}{2} + \frac{y}{2} + 1\right)dx\right)dy
=\ldots=\int\limits_{-3}^{0}\left( - \frac{221 y^{2}}{300} - \frac{493 y}{150} - \frac{323}{100} \right)dy + \int\limits_{0}^{2}\left( - \frac{1003 y^{2}}{400} + \frac{663 y}{100} - \frac{323}{100} \right)dy = \ldots = - \frac{17}{12}$ \\
***********************************************************************************************
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 + 2 y - 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,0), B=(2,0), C=(3,-1)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{2}\left(\int\limits_{- \frac{x}{3}}^{0}\left(- 2 x + 2 y - 1\right)dy\right)dx + \int\limits_{2}^{3}\left(\int\limits_{- \frac{x}{3}}^{2 - x}\left(- 2 x + 2 y - 1\right)dy\right)dx =\ldots=\int\limits_{0}^{2}\left( - \frac{7 x^{2}}{9} - \frac{x}{3} \right)dx + \int\limits_{2}^{3}\left( \frac{20 x^{2}}{9} - \frac{22 x}{3} + 2 \right)dx = \ldots = -5$$
Względem Oy:
$$\int\limits_{-1}^{0}\left(\int\limits_{- 3 y}^{2 - y}\left(- 2 x + 2 y - 1\right)dx\right)dy =\ldots=\int\limits_{-1}^{0}\left(12 y^{2} + 6 y - 6\right)dy = \ldots = -5$$
***********************************************************************************************
** Zadanie 2 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + 2 y + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,0), B=(-2,0), C=(0,3)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-2}^{0}\left(\int\limits_{0}^{\frac{3 x}{2} + 3}\left(- 2 x + 2 y + 2\right)dy\right)dx =\ldots=\int\limits_{-2}^{0}\left(- \frac{3 x^{2}}{4} + 6 x + 15\right)dx = \ldots = 16$$
Względem Oy:
$$\int\limits_{0}^{3}\left(\int\limits_{\frac{2 y}{3} - 2}^{0}\left(- 2 x + 2 y + 2\right)dx\right)dy =\ldots=\int\limits_{0}^{3}\left(8 - \frac{8 y^{2}}{9}\right)dy = \ldots = 16$$
***********************************************************************************************
** Zadanie 3 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x - 3 y + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,-3), B=(0,0), C=(2,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{2}\left(\int\limits_{\frac{3 x}{2} - 3}^{0}\left(- x - 3 y + 2\right)dy\right)dx =\ldots=\int\limits_{0}^{2}\left(\frac{39 x^{2}}{8} - \frac{39 x}{2} + \frac{39}{2}\right)dx = \ldots = 13$$
Względem Oy:
$$\int\limits_{-3}^{0}\left(\int\limits_{0}^{\frac{2 y}{3} + 2}\left(- x - 3 y + 2\right)dx\right)dy =\ldots=\int\limits_{-3}^{0}\left(- \frac{20 y^{2}}{9} - 6 y + 2\right)dy = \ldots = 13$$
***********************************************************************************************
** Zadanie 4 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 3 x + 2 y + 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(0,3), B=(2,1), C=(1,4)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{1}\left(\int\limits_{3 - x}^{x + 3}\left(- 3 x + 2 y + 1\right)dy\right)dx + \int\limits_{1}^{2}\left(\int\limits_{3 - x}^{7 - 3 x}\left(- 3 x + 2 y + 1\right)dy\right)dx =\ldots=\int\limits_{0}^{1}\left( - 6 x^{2} + 14 x \right)dx + \int\limits_{1}^{2}\left( 14 x^{2} - 50 x + 44 \right)dx = \ldots = \frac{20}{3}$$
Względem Oy:
$$\int\limits_{1}^{3}\left(\int\limits_{3 - y}^{\frac{7}{3} - \frac{y}{3}}\left(- 3 x + 2 y + 1\right)dx\right)dy + \int\limits_{3}^{4}\left(\int\limits_{y - 3}^{\frac{7}{3} - \frac{y}{3}}\left(- 3 x + 2 y + 1\right)dx\right)dy =\ldots=\int\limits_{1}^{3}\left( \frac{8 y^{2}}{3} - \frac{22 y}{3} + \frac{14}{3} \right)dy + \int\limits_{3}^{4}\left( - \frac{4 y^{2}}{3} + \frac{8 y}{3} + \frac{32}{3} \right)dy = \ldots = \frac{20}{3}$$
***********************************************************************************************
** Zadanie 5 **********************************************************************************
Obliczyć $$\iint\limits_D\left(x - 3 y + 1\right)dxdy$$ gdzie D - trójkąt ABC dla A=(2,0), B=(0,1), C=(0,-3)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{2}\left(\int\limits_{\frac{3 x}{2} - 3}^{1 - \frac{x}{2}}\left(x - 3 y + 1\right)dy\right)dx =\ldots=\int\limits_{0}^{2}\left(x^{2} - 10 x + 16\right)dx = \ldots = \frac{44}{3}$$
Względem Oy:
$$\int\limits_{-3}^{0}\left(\int\limits_{0}^{\frac{2 y}{3} + 2}\left(x - 3 y + 1\right)dx\right)dy + \int\limits_{0}^{1}\left(\int\limits_{0}^{2 - 2 y}\left(x - 3 y + 1\right)dx\right)dy =\ldots=\int\limits_{-3}^{0}\left( - \frac{16 y^{2}}{9} - 4 y + 4 \right)dy + \int\limits_{0}^{1}\left( 8 y^{2} - 12 y + 4 \right)dy = \ldots = \frac{44}{3}$$
***********************************************************************************************
** Zadanie 6 **********************************************************************************
Obliczyć $$\iint\limits_D\left(x + \frac{y}{2} + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-2,1), B=(-2,0), C=(1,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-2}^{1}\left(\int\limits_{0}^{\frac{1}{3} - \frac{x}{3}}\left(x + \frac{y}{2} + 2\right)dy\right)dx =\ldots=\int\limits_{-2}^{1}\left(- \frac{11 x^{2}}{36} - \frac{7 x}{18} + \frac{25}{36}\right)dx = \ldots = \frac{7}{4}$$
Względem Oy:
$$\int\limits_{0}^{1}\left(\int\limits_{-2}^{1 - 3 y}\left(x + \frac{y}{2} + 2\right)dx\right)dy =\ldots=\int\limits_{0}^{1}\left(3 y^{2} - \frac{15 y}{2} + \frac{9}{2}\right)dy = \ldots = \frac{7}{4}$$
***********************************************************************************************
** Zadanie 7 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{3 x}{2} + \frac{y}{2}\right)dxdy$$ gdzie D - trójkąt ABC dla A=(2,-1), B=(3,0), C=(0,3)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{2}\left(\int\limits_{3 - 2 x}^{3 - x}\left(\frac{3 x}{2} + \frac{y}{2}\right)dy\right)dx + \int\limits_{2}^{3}\left(\int\limits_{x - 3}^{3 - x}\left(\frac{3 x}{2} + \frac{y}{2}\right)dy\right)dx =\ldots=\int\limits_{0}^{2}\left( \frac{3 x^{2}}{4} + \frac{3 x}{2} \right)dx + \int\limits_{2}^{3}\left( - 3 x^{2} + 9 x \right)dx = \ldots = \frac{17}{2}$$
Względem Oy:
$$\int\limits_{-1}^{0}\left(\int\limits_{\frac{3}{2} - \frac{y}{2}}^{y + 3}\left(\frac{3 x}{2} + \frac{y}{2}\right)dx\right)dy + \int\limits_{0}^{3}\left(\int\limits_{\frac{3}{2} - \frac{y}{2}}^{3 - y}\left(\frac{3 x}{2} + \frac{y}{2}\right)dx\right)dy =\ldots=\int\limits_{-1}^{0}\left( \frac{21 y^{2}}{16} + \frac{51 y}{8} + \frac{81}{16} \right)dy + \int\limits_{0}^{3}\left( \frac{5 y^{2}}{16} - \frac{21 y}{8} + \frac{81}{16} \right)dy = \ldots = \frac{17}{2}$$
***********************************************************************************************
** Zadanie 8 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + y + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-1,0), B=(0,0), C=(0,4)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-1}^{0}\left(\int\limits_{0}^{4 x + 4}\left(- 2 x + y + 2\right)dy\right)dx =\ldots=\int\limits_{-1}^{0}\left(16 x + 16\right)dx = \ldots = 8$$
Względem Oy:
$$\int\limits_{0}^{4}\left(\int\limits_{\frac{y}{4} - 1}^{0}\left(- 2 x + y + 2\right)dx\right)dy =\ldots=\int\limits_{0}^{4}\left(3 - \frac{3 y^{2}}{16}\right)dy = \ldots = 8$$
***********************************************************************************************
** Zadanie 9 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 3 x + 2 y + 2\right)dxdy$$ gdzie D - trójkąt ABC dla A=(-3,0), B=(0,0), C=(0,-2)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-3}^{0}\left(\int\limits_{- \frac{2 x}{3} - 2}^{0}\left(- 3 x + 2 y + 2\right)dy\right)dx =\ldots=\int\limits_{-3}^{0}\left(- \frac{22 x^{2}}{9} - \frac{22 x}{3}\right)dx = \ldots = 11$$
Względem Oy:
$$\int\limits_{-2}^{0}\left(\int\limits_{- \frac{3 y}{2} - 3}^{0}\left(- 3 x + 2 y + 2\right)dx\right)dy =\ldots=\int\limits_{-2}^{0}\left(\frac{51 y^{2}}{8} + \frac{45 y}{2} + \frac{39}{2}\right)dy = \ldots = 11$$
***********************************************************************************************
** Zadanie 10 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{x}{2} - y\right)dxdy$$ gdzie D - trójkąt ABC dla A=(1,0), B=(4,-1), C=(4,0)
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{1}^{4}\left(\int\limits_{\frac{1}{3} - \frac{x}{3}}^{0}\left(\frac{x}{2} - y\right)dy\right)dx =\ldots=\int\limits_{1}^{4}\left(\frac{2 x^{2}}{9} - \frac{5 x}{18} + \frac{1}{18}\right)dx = \ldots = \frac{11}{4}$$
Względem Oy:
$$\int\limits_{-1}^{0}\left(\int\limits_{1 - 3 y}^{4}\left(\frac{x}{2} - y\right)dx\right)dy =\ldots=\int\limits_{-1}^{0}\left(- \frac{21 y^{2}}{4} - \frac{3 y}{2} + \frac{15}{4}\right)dy = \ldots = \frac{11}{4}$$
***********************************************************************************************
Po obszarze ograniczonym krzywymi
zadanie = gz.calka_podwojna(typ=2, nr_zadania=1)
zadanie
('Obliczyć $$\\iint\\limits_D\\left(- 2 y\\right)dx\\,dy$$ gdzie $D$ - obszar ograniczony krzywymi $y=x^{2} - x$, oraz $y=- 2 x$',
'\\includegraphics[width=45mm]{../pics/calka_podwojna1x}\n\t\\includegraphics[width=45mm]{../pics/calka_podwojna1y}\\\\\n\tWzględem $Ox:$ $\\int\\limits_{-1}^{0}\\left(\\int\\limits_{x^{2} - x}^{- 2 x}\\left(- 2 y\\right)dy\\right)dx\n\t=\\ldots=\\int\\limits_{-1}^{0}\\left(x^{4} - 2 x^{3} - 3 x^{2}\\right)dx = \\ldots = - \\frac{3}{10}$ \\\\')
Show code cell source
print("\033[34m** Zadanie **" + '*' * 81 + '\033[0m')
print(zadanie[0])
print("\033[34m*\033[0m" * 95)
print("\033[32m** Rozwiązanie **" + '*' * 78 + '\033[0m')
print(zadanie[1])
print("\033[32m*\033[0m" * 95)
** Zadanie ***********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 y\right)dx\,dy$$ gdzie $D$ - obszar ograniczony krzywymi $y=x^{2} - x$, oraz $y=- 2 x$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
\includegraphics[width=45mm]{../pics/calka_podwojna1x}
\includegraphics[width=45mm]{../pics/calka_podwojna1y}\\
Względem $Ox:$ $\int\limits_{-1}^{0}\left(\int\limits_{x^{2} - x}^{- 2 x}\left(- 2 y\right)dy\right)dx
=\ldots=\int\limits_{-1}^{0}\left(x^{4} - 2 x^{3} - 3 x^{2}\right)dx = \ldots = - \frac{3}{10}$ \\
***********************************************************************************************
Show code cell source
for i in range(ile_zadan_przykladowych, 2 * ile_zadan_przykladowych + 1):
zadanie = gz.calka_podwojna(typ=2, nr_zadania=i)
print(f"\033[34m** Zadanie {i} **" + '*' * 80 + '\033[0m')
display(Markdown(
zadanie[0].split('krzywymi')[0].replace('\\,', '').replace('$', '*', ).replace('**', '$$', 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 $$\iint\limits_D\left(- x + y - 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=2 y^{2} - 3 y\quad \text{ oraz } \quad x=3 y$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{0}^{3}\left(\int\limits_{2 y^{2} - 3 y}^{3 y}\left(- x + y - 1\right)dx\right)dy =\ldots=\int\limits_{0}^{3}\left(2 y^{4} - 8 y^{3} + 8 y^{2} - 6 y\right)dy = \ldots = - \frac{99}{5}$$
***********************************************************************************************
** Zadanie 11 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x - 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=y^{2} + 3 y - 1\quad \text{ oraz } \quad x=3 y$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-1}^{1}\left(\int\limits_{y^{2} + 3 y - 1}^{3 y}\left(- x - 1\right)dx\right)dy =\ldots=\int\limits_{-1}^{1}\left(\frac{y^{4}}{2} + 3 y^{3} - 3 y - \frac{1}{2}\right)dy = \ldots = - \frac{4}{5}$$
***********************************************************************************************
** Zadanie 12 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 2 x + y + 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=y^{2} - 1\quad \text{ oraz } \quad x=- 3 y - 3$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-2}^{-1}\left(\int\limits_{y^{2} - 1}^{- 3 y - 3}\left(- 2 x + y + 1\right)dx\right)dy =\ldots=\int\limits_{-2}^{-1}\left(y^{4} - y^{3} - 15 y^{2} - 23 y - 10\right)dy = \ldots = - \frac{11}{20}$$
***********************************************************************************************
** Zadanie 13 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{3 x}{2} + 2 y + 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=y^{2} + 3 y - 3\quad \text{ oraz } \quad x=y - 3$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-2}^{0}\left(\int\limits_{y^{2} + 3 y - 3}^{y - 3}\left(\frac{3 x}{2} + 2 y + 1\right)dx\right)dy =\ldots=\int\limits_{-2}^{0}\left(- \frac{3 y^{4}}{4} - \frac{13 y^{3}}{2} - \frac{13 y^{2}}{2} + 7 y\right)dy = \ldots = - \frac{152}{15}$$
***********************************************************************************************
** Zadanie 14 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- 3 x - 2 y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=\frac{y^{2}}{2}\quad \text{ oraz } \quad x=3 y$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{0}^{6}\left(\int\limits_{\frac{y^{2}}{2}}^{3 y}\left(- 3 x - 2 y\right)dx\right)dy =\ldots=\int\limits_{0}^{6}\left(\frac{3 y^{4}}{8} + y^{3} - \frac{39 y^{2}}{2}\right)dy = \ldots = - \frac{2484}{5}$$
***********************************************************************************************
** Zadanie 15 **********************************************************************************
Obliczyć $$\iint\limits_D\left(y\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=\frac{x^{2}}{2}\quad \text{ oraz } \quad y=- 2 x$$
***********************************************************************************************
** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{-4}^{0}\left(\int\limits_{\frac{x^{2}}{2}}^{- 2 x}\left(y\right)dy\right)dx =\ldots=\int\limits_{-4}^{0}\left(- \frac{x^{4}}{8} + 2 x^{2}\right)dx = \ldots = \frac{256}{15}$$
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** Zadanie 16 **********************************************************************************
Obliczyć $$\iint\limits_D\left(- x\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=y^{2} - 3 y - 3\quad \text{ oraz } \quad x=2 y + 3$$
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** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-1}^{6}\left(\int\limits_{y^{2} - 3 y - 3}^{2 y + 3}\left(- x\right)dx\right)dy =\ldots=\int\limits_{-1}^{6}\left(\frac{y^{4}}{2} - 3 y^{3} - \frac{y^{2}}{2} + 3 y\right)dy = \ldots = - \frac{10633}{60}$$
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** Zadanie 17 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{3 x}{2} + 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=- 3 y^{2} + y\quad \text{ oraz } \quad x=- 2 y$$
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** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{0}^{1}\left(\int\limits_{- 2 y}^{- 3 y^{2} + y}\left(\frac{3 x}{2} + 1\right)dx\right)dy =\ldots=\int\limits_{0}^{1}\left(\frac{27 y^{4}}{4} - \frac{9 y^{3}}{2} - \frac{21 y^{2}}{4} + 3 y\right)dy = \ldots = - \frac{1}{40}$$
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** Zadanie 18 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 x + \frac{y}{2} + 2\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=\frac{y^{2}}{2} + \frac{y}{2} + 1\quad \text{ oraz } \quad x=2 y + 3$$
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** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-1}^{4}\left(\int\limits_{\frac{y^{2}}{2} + \frac{y}{2} + 1}^{2 y + 3}\left(2 x + \frac{y}{2} + 2\right)dx\right)dy =\ldots=\int\limits_{-1}^{4}\left(- \frac{y^{4}}{4} - \frac{3 y^{3}}{4} + \frac{5 y^{2}}{2} + 15 y + 12\right)dy = \ldots = \frac{6125}{48}$$
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** Zadanie 19 **********************************************************************************
Obliczyć $$\iint\limits_D\left(\frac{x}{2}\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$y=x^{2}\quad \text{ oraz } \quad y=3 x$$
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** Rozwiązanie ********************************************************************************
Względem Ox:
$$\int\limits_{0}^{3}\left(\int\limits_{x^{2}}^{3 x}\left(\frac{x}{2}\right)dy\right)dx =\ldots=\int\limits_{0}^{3}\left(- \frac{x^{3}}{2} + \frac{3 x^{2}}{2}\right)dx = \ldots = \frac{27}{8}$$
Względem Oy:
$$\int\limits_{0}^{9}\left(\int\limits_{\frac{y}{3}}^{\sqrt{y}}\left(\frac{x}{2}\right)dx\right)dy =\ldots=\int\limits_{0}^{9}\left(- \frac{y^{2}}{36} + \frac{y}{4}\right)dy = \ldots = \frac{27}{8}$$
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** Zadanie 20 **********************************************************************************
Obliczyć $$\iint\limits_D\left(2 x + 1\right)dxdy$$ gdzie D - obszar ograniczony krzywymi
$$x=y^{2} - 2\quad \text{ oraz } \quad x=2 y + 1$$
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** Rozwiązanie ********************************************************************************
Względem Oy:
$$\int\limits_{-1}^{3}\left(\int\limits_{y^{2} - 2}^{2 y + 1}\left(2 x + 1\right)dx\right)dy =\ldots=\int\limits_{-1}^{3}\left(- y^{4} + 7 y^{2} + 6 y\right)dy = \ldots = \frac{608}{15}$$
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