Решение .
Вычислить двoйной интеграл по области D .
[tex]\displaystyle \bf D:\ \{\ y=\sqrt{x}\ ,\ \ y=-x^2\ ,\ \ x=1\ \}[/tex]
Точки пересечения линий :
[tex]\bf \sqrt{x}=-x^2\ \ ,\ \ x^2+\sqrt{x}=0\ \ ,\ \ \sqrt{x}\, (x\sqrt{x}+1)=0\ \ ,\ \ x=0\\\\y=\sqrt{x}\ ,\ \ x=1\ \ ,\ \ y(1)=\sqrt1=1\\\\y=-x^2\ \ ,\ \ y(1)=-1^2=-1[/tex]
[tex]\displaystyle \bf \iint \limits_{D}\, (3x-8y)\, dx\, dy=\int \limits _0^1\, dx\int \limits _{-x^2}^{\sqrt{x}}(3x-8y)\, dy=\int \limits _0^1\, dx\Big(3xy-4y^2\Big)\Big|_{-x^2}^{\sqrt{x}}=\\\\\\=\int \limits _0^1\, \Big(3x\sqrt{x}-4x-3x(-x^2)+4(-x^2)^2\Big)\, dx=\\\\\\=\int \limits _0^1\, \Big(3x^{\frac{3}{2}}-4x+3x^3+4x^4\Big)\, dx=\Big(\frac{3x^{\frac{5}{2}}}{5/2}-2x^2+\frac{3x^4}{4}+\frac{4x^5}{5}\Big)\Big|_0^1=[/tex]
[tex]\bf \displaystyle =\frac{6}{2}-2+\frac{3}{4}+\frac{4}{5}=\frac{60-40+15+16}{20}=\frac{51}{40}=1,275[/tex]
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Решение .
Вычислить двoйной интеграл по области D .
[tex]\displaystyle \bf D:\ \{\ y=\sqrt{x}\ ,\ \ y=-x^2\ ,\ \ x=1\ \}[/tex]
Точки пересечения линий :
[tex]\bf \sqrt{x}=-x^2\ \ ,\ \ x^2+\sqrt{x}=0\ \ ,\ \ \sqrt{x}\, (x\sqrt{x}+1)=0\ \ ,\ \ x=0\\\\y=\sqrt{x}\ ,\ \ x=1\ \ ,\ \ y(1)=\sqrt1=1\\\\y=-x^2\ \ ,\ \ y(1)=-1^2=-1[/tex]
[tex]\displaystyle \bf \iint \limits_{D}\, (3x-8y)\, dx\, dy=\int \limits _0^1\, dx\int \limits _{-x^2}^{\sqrt{x}}(3x-8y)\, dy=\int \limits _0^1\, dx\Big(3xy-4y^2\Big)\Big|_{-x^2}^{\sqrt{x}}=\\\\\\=\int \limits _0^1\, \Big(3x\sqrt{x}-4x-3x(-x^2)+4(-x^2)^2\Big)\, dx=\\\\\\=\int \limits _0^1\, \Big(3x^{\frac{3}{2}}-4x+3x^3+4x^4\Big)\, dx=\Big(\frac{3x^{\frac{5}{2}}}{5/2}-2x^2+\frac{3x^4}{4}+\frac{4x^5}{5}\Big)\Big|_0^1=[/tex]
[tex]\bf \displaystyle =\frac{6}{2}-2+\frac{3}{4}+\frac{4}{5}=\frac{60-40+15+16}{20}=\frac{51}{40}=1,275[/tex]