Задача 64000 4и5. Вычислить криволинейный интеграл...
Условие
Решение
[m]dl=\sqrt{1+y`(x)}dx[/m]
[m]y`(x)=(tgx)`=\frac{1}{cos^2x}[/m]
тогда
[m]dl=\sqrt{1+(\frac{1}{cos^2x})^2}dx=\sqrt{\frac{cos^4x+1}{cos^4x}}dx=\frac{\sqrt{cos^4x+1}}{cos^2x}dx[/m]
[m] ∫_{L} \sqrt{1+cos^4x} dl= ∫_{0} ^{\frac{π}{4}}\sqrt{1+cos^4x}\cdot =\frac{\sqrt{cos^4x+1}}{cos^2x}dx=∫_{0} ^{\frac{π}{4}}\frac{1+cos^4x}{cos^2x}dx=[/m]
[m]=∫_{0} ^{\frac{π}{4}}\frac{1}{cos^2x}dx+∫_{0} ^{\frac{π}{4}}\frac{cos^4x}{cos^2x}dx=[/m]
[m]=∫_{0} ^{\frac{π}{4}}\frac{1}{cos^2x}dx+∫_{0} ^{\frac{π}{4}}cos^2xdx=[/m]
[m]=∫_{0} ^{\frac{π}{4}}\frac{1}{cos^2x}dx+∫_{0} ^{\frac{π}{4}}\frac{1+cos2x}{2}dx=[/m]
[m]=(tgx +\frac{1}{2}x+\frac{1}{2}\cdot \frac{1}{2}sin2x)|_{0} ^{\frac{π}{4}}=tg\frac{π}{4} +\frac{1}{2}\frac{π}{4}+\frac{1}{2}\cdot \frac{1}{2}sin(2\cdot \frac{π}{4})=1+\frac{π}{2}+\frac{1}{4}[/m]
5.
x^2+y^2=9 ⇒ y=-sqrt(9-x^2) или y=+sqrt(9-x^2) - две полуокружности, в нижней полуплоскости и верхней.
(см. рис)
dy=y`dx
[m]dy=(-\sqrt{9-x^2} )`dx=-\frac{1}{2\sqrt{9-x^2}}\cdot (9-x^2)`dx=-\frac{1}{2\sqrt{9-x^2}}\cdot (-2x)dx=\frac{1}{\sqrt{9-x^2}}dx[/m]
[m]dy=(\sqrt{9-x^2} )`dx=\frac{1}{2\sqrt{9-x^2}}\cdot (9-x^2)`dx=\frac{1}{2\sqrt{9-x^2}}\cdot (-2x)dx=\frac{1}{\sqrt{9-x^2}}dx[/m]
И подставляем в выражения под интегралом.
1) по кривой y=-sqrt(9-x^2)
[m] ∫_{L_{1}} (y^2-y)dx+(2xy+x) dy= ∫_{-3} ^{3}(9-x^2-(-\sqrt{9-x^2})dx+(2x(-\sqrt{9-x^2})+x)\cdot \frac{1}{\sqrt{9-x^2}}dx=[/m]
[m]=∫_{-3} ^{3}(9-x^2+\sqrt{9-x^2}-2x+\frac{2x}{\sqrt{9-x^2}})dx=(9x-\frac{x^3}{3}+\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}arcsin\frac{x}{3} -x^2-2\sqrt{9-x^2})|_{3} ^{-3}[/m]
Получили определенный интеграл
Считайте
2)по кривой y=sqrt(9-x^2)
[m] ∫_{L_{2}} (y^2-y)dx+(2xy+x) dy= ∫_{3} ^{-3}(9-x^2-(\sqrt{9-x^2})dx+(2x(\sqrt{9-x^2})+x)\cdot (-\frac{1}{\sqrt{9-x^2}})dx=[/m]
[m]=∫_{3} ^{-3}(9-x^2-\sqrt{9-x^2}-2x-\frac{2x}{\sqrt{9-x^2}})dx=(9x-\frac{x^3}{3}-\frac{x}{2}\sqrt{9-x^2}-\frac{9}{2}arcsin\frac{x}{3} -x^2+2\sqrt{9-x^2})|_{3} ^{-3}[/m]
Получили определенный интеграл
Считайте
Тогда
[m] ∫_{L} (y^2-y)dx+(2xy+x) dy=∫_{L_{1}} (y^2-y)dx+(2xy+x) dy+∫_{L_{2}} (y^2-y)dx+(2xy+x) dy=[/m]
Второй способ
Параметрическое задание кривой
x=3sint
y=3cost
dx=3costdt
dy=-3sintdt
[m] ∫_{L} (y^2-y)dx+(2xy+x) dy= ∫_{-π} ^{π}((3cost)^2-3cost)\cdot 3 cost+(2\cdot 3sint\cdot 3cost+2\cdot 3sint)\cdot (-3sint)dt=[/m]
[m] = ∫_{-π} ^{π}(27 cos^2t\cdot cost -9cos^2t+27sin^2t \cdot cost-18sin^2t)dt=[/m]
[m] = ∫_{-π} ^{π}(27 (1-sin^2t)cost-9cos^2t+27sin^2t \cdot cost-18sin^2t)dt=[/m]
[m] = ∫_{-π} ^{π}(27cost -27sin^2t\cdot cost-9cos^2t+27sin^2t \cdot cost-18sin^2t)dt=[/m]
[m] = ∫_{-π} ^{π}(27cost -9cos^2t-9sin^2t-9 sin^2t)dt=[/m]
[m] = ∫_{-π} ^{π}(27cost -9-9 \cdot \frac{1-cos2t}{2})dt=[/m]
