Задача 61725 Вычислить площадь фигуры, ограниченной...
Условие
Решение
[m]8sint=4[/m] ⇒ [m]sint=\frac{1}{2}[/m] ⇒[m] t_{1}=\frac{π}{6}; t_{2}=\frac{5π}{6}[/m]
значит.
[m]S= ∫ _{\frac{5π}{6}} ^{\frac{π}{6}} 8sint \cdot (3cost)`dt-∫ _{\frac{5π}{6}} ^{\frac{π}{6}}4dt=∫_{\frac{5π}{6}} ^{\frac{π}{6}}8sint \cdot (-3sint)dt-∫_{\frac{5π}{6}} ^{\frac{π}{6}}4dt= [/m]
[m]S=24∫ _{\frac{π}{6}}^{\frac{5π}{6}} sin^2t dt+∫_{\frac{π}{6}} ^{\frac{5π}{6}}4dt= [/m]
[m]=24∫ _{\frac{π}{6}}^{\frac{5π}{6}} \frac{1-cos2t}{2} dt+∫_{\frac{π}{6}} ^{\frac{5π}{6}}4dt= [/m]
[m]=12∫ _{\frac{π}{6}}^{\frac{5π}{6}}(1-cos2t)+ dt∫_{\frac{π}{6}} ^{\frac{5π}{6}}4dt= [/m]
[m]=(12(t-\frac{1}{2}sin2t))| _{\frac{π}{6}}^{\frac{5π}{6}}+4(t)|_{\frac{π}{6}}^{\frac{5π}{6}}=12((\frac{5π}{6}-(\frac{π}{6}))-6sin (2\cdot\frac{5π}{6})+6sin(2\cdot \frac{π}{6})+4(\frac{5π}{6}-\frac{π}{6})=8π-6sin\frac{5π}{3}+6sin\frac{π}{3}+\frac{8π}{3}=[/m]
[m]=\frac{32π}{3}-6(-\frac{\sqrt{3}}{2})+6(\frac{\sqrt{3}}{2})=\frac{32π}{3}+6\sqrt{3}[/m]
