Respuesta :
Parameterize [tex]S[/tex] by
[tex]\vec r(u,v)=u\,\vec\imath+4\cos v\,\vec\jmath+4\sin v\,\vec k[/tex]
with [tex]0\le u\le 3[/tex] and [tex]0\le v\le\dfrac\pi2[/tex]. The normal vector to [tex]S[/tex] has magnitude
[tex]\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=4[/tex]
so that the surface integral is
[tex]\displaystyle\iint_S(z+x^2y)\,\mathrm dS=4\int_0^{\pi/2}\int_0^3(4\sin v+4u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{192}[/tex]
The value of the surface integral which is part of the cylinder that lies between the planes x = 0 and x = 3 in the first octant is 192
The process of arriving at the above value is as follows:
Known parameters:
Question: The surface integral equation appears to have some parts left out as follows;
[tex]\displaystyle \int\int\limits_s {(z + x^2 \cdot y)} \, ds[/tex]
The equation of the cylinder is y² + z² = 16
The planes between which the cylinder lies = x = 0, and x = 3
The required parameter:
Evaluate the surface integral
Method:
We have;
z² = 16 - y²
∴ z = √(16 - y²)
Therefore;
z is a function of x, and y
z = g(x, y) = √(16 - y²)
g'ₓ = d(√(16 - y²))/dx = 0
[tex]g'_y = \dfrac{d(\sqrt{16 - y^2} }{dy} = \left(\dfrac{1}{2} \right) \times \dfrac{-2\cdot y}{\sqrt{16 - y^2}} = - \dfrac{ y}{\sqrt{16 - y^2}}[/tex]
The formula for surface integral is presented as follows;
[tex]\displaystyle \int \int\limits_S {f(x, y, z)} \, dS =\mathbf{ \int\int\limits_D {f(x, y, g(x, y))} \times \sqrt{\left(\frac{\partial g}{\partial x} \right)^2 + \left(\frac{\partial g}{\partial y} \right)^2 + 1} \, dA}[/tex]
Substituting gives;
[tex]\displaystyle \int\int\limits_D {\left(\sqrt{16-y^2} + x^2 \cdot y\right) } \times \sqrt{\left(1 + \dfrac{y^2}{16 - y^2}} \right)}\, dA[/tex]
[tex]\displaystyle \int\int\limits_D {\left(\sqrt{16-y^2} + x^2 \cdot y\right) } \times \sqrt{\left( \dfrac{16 - y^2 + y^2}{16 - y^2}} \right)}\, dA[/tex]
[tex]\displaystyle \int\int\limits_D {\left(\sqrt{16-y^2} + x^2 \cdot y\right) } \times \left( \dfrac{4}{\sqrt{ 16 - y^2}}} \right)}\, dA = \int\int\limits_D {\left(1+ \dfrac{x^2 \cdot y \cdot 4}{\sqrt{ 16 - y^2}}} \right)}\, dA[/tex]
[tex]\left \displaystyle \int_0^3\int\limits_{0}^4 {\left(1+ \dfrac{x^2 \cdot y \cdot 4}{\sqrt{ 16 - y^2}}} \right)}\, dy \cdot dx = \int_0^3 4\cdot y\right |_0^4 \, dx + \displaystyle \left \int\limits_0^3 {\left( {-4\cdot x^2 \cdot \sqrt{ 16 - y^2}}} \right)}\, \right |_0^4dx[/tex]
Given that we have;
[tex]\mathbf{\left {\left( {-4\cdot \sqrt{ 16 - y^2}}} \right)}\, \right |_0^4 }= -4 \times -4 = \mathbf{16}[/tex]
Therefore
[tex]\left \displaystyle \int_0^3 16 \, dx - \int\limits_0^3 {\left( {4\cdot x^2 \cdot \sqrt{ 16 - y^2}}} \right)}\, \right |_0^4dx = 48 +16\times \displaystyle \int_0^3 x^2 \, dx[/tex]
[tex]48 +16\times \displaystyle \int_0^3 x^2 \, dx = 48 + 16 \times \dfrac{27}{3} =192[/tex]
The value of the surface integral = 192
Learn more about surface integral here:
https://brainly.com/question/10034572
