Respuesta :
By Green's theorem,
[tex]\displaystyle\int_C(7(x^2-y)\,\vec\imath+3(y^2+x)\,\vec\jmath)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial3(y^2+x)}{\partial x}-\frac{\partial7(x^2-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]\displaystyle=10\iint_D\mathrm dx\,\mathrm dy[/tex]
where [tex]D[/tex] is the region bounded by the closed curve [tex]C[/tex]. The remaining integral is 10 times the area of [tex]D[/tex].
Since [tex]D[/tex] is a circle in both cases, and we're given the equations for them right away, it's just a matter of determining the radius of each one and plugging it into the well-known formula for the area of a circle with radius [tex]r[/tex], [tex]\pi r^2[/tex].
(a) [tex]C[/tex] is a circle with radius 3, so the line integral is [tex]10\pi(3^2)=\boxed{90\pi}[/tex].
(b) [tex]C[/tex] is a circle with radius [tex]R[/tex], so the line integral is [tex]\boxed{10\pi R^2}[/tex].
- A circle C with the radius [tex]3[/tex] is [tex]90\pi[/tex].
- A circle C with the radius [tex]R[/tex] is [tex]10\pi R^2[/tex].
Let ,
First calculate the value for given equation by using green's theorem,
Since, the formula for greens theorem is:
- [tex]\int\ CFds=\int\ \int\ CurlFkdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ CurlFkdA[/tex]...(1)
The given equation is,
[tex]\int\ C [7(x^2-y)i +3(y^2+x)j]dr[/tex]
Here,
[tex]F=[7(x^2-y)i +3(y^2+x)j][/tex]
Now to calculate the value of [tex]Curl F[/tex],
[tex]Curl F=\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial k} \\7(x^2-y)&(y^2+x)&0\end{array}\right] \\\\CurlF=[\frac{\partial}{ \partial y} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (y^2+x)-\frac{\partial}{\partial y} (7x^2-y)]\\\\Curl F=0+0+10k\\\\CurlF=10k[/tex]
Substitute in equation (1),
[tex]\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10k *kdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10dA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=10\int\ \int\ dA[/tex]
The remaining integral is [tex]10[/tex] times the area of region .
The general equation is,
[tex]x^2+y^2=r^2[/tex]
The area of a circle is [tex]\pi r^2[/tex] .
Hence, area of region of circle is [tex]10\pi r^2[/tex].
Now,
(a) The given equation is [tex](x-2)^2+(y-3)^2=3^2[/tex],
C is a circle with radius 3, so the line integral is
[tex]10\pi (3)^2=90\pi[/tex] .
(b) The given equation is [tex](x-a)^2+(y-b)^2=R^2[/tex],
C is a circle with radius [tex]R[/tex] , so the line integral is,
[tex]10\pi (R)^2=10\pi R^2[/tex].
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