Respuesta :
Write the line integral as
[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy[/tex]
In other words, [tex]\mathbf f(x,y)=P(x,y)\,\mathbf i+Q(x,y)\,\mathbf j[/tex], and [tex]\mathrm d\mathbf r=\mathrm dx\,\mathbf i+\mathrm dy\,\mathbf j[/tex].
The curve [tex]\mathcal C[/tex] is closed, so we can apply Green's theorem:
[tex]\displastyle\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy=\iint_{\mathcal D}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}\bigg((4x-3y^2)-(3x^2)\bigg)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}(4x-3x^2-3y^2)\,\mathrm dx\,\mathrm dy[/tex]
Convert to polar coordinates to write this as
[tex]\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r\cos\theta-3r^2\cos^2\theta-3r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r^2\cos\theta-3r^3)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\dfrac{16\sqrt2}3-3\pi[/tex]
[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy[/tex]
In other words, [tex]\mathbf f(x,y)=P(x,y)\,\mathbf i+Q(x,y)\,\mathbf j[/tex], and [tex]\mathrm d\mathbf r=\mathrm dx\,\mathbf i+\mathrm dy\,\mathbf j[/tex].
The curve [tex]\mathcal C[/tex] is closed, so we can apply Green's theorem:
[tex]\displastyle\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy=\iint_{\mathcal D}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}\bigg((4x-3y^2)-(3x^2)\bigg)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}(4x-3x^2-3y^2)\,\mathrm dx\,\mathrm dy[/tex]
Convert to polar coordinates to write this as
[tex]\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r\cos\theta-3r^2\cos^2\theta-3r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r^2\cos\theta-3r^3)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\dfrac{16\sqrt2}3-3\pi[/tex]
The line integral of curve f=(3x2y)i+(2x2−3xy2)j is = 16√2/3 - 3π.
What is line integral?
A line integral is an integral where the function is determined as the integration along the curve in a coordinate system.
[tex]\rm \int\limits f(x, y). dr = \int\limits Pdx +Qdy[/tex]
[tex]\rm \int\limits f(x, y). dr = P(x, y)i +Q(x, y)j\\dr = idx + jdy[/tex]
The curve is closed, so we can apply Green's theorem:
[tex]\rm \int\limits Pdx +Qdy = \int \int(dQ/dx - dP/dy) dx dy[/tex]
[tex]\rm = \int \int[(4x - 3y^{2} )- 3x^{2} ] dx dy\\\rm = \int \int[4x - 3y^{2} - 3x^{2} ] dx dy\\[/tex]
Converting to polar coordinate
[tex]\rm \int\limits^\frac{\pi }{4} _0 \int\limits^2_0(4r cos \theta - 3r^{2} cos^{2} \theta -3r^{2} sin^{2} \theta) rdr d\theta[/tex]
[tex]\rm \int\limits^\frac{\pi }{4} _0 \int\limits^2_0( 4r^{2} cos\theta -3r^{3} ) rdr d\theta[/tex]
= 16√2/3 - 3π
Therefore, the line integral of curve f=(3x2y)i+(2x2−3xy2)j is = 16√2/3 - 3π.
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