Respuesta :
See attached image for a plot of the region. Red represents the region of interest, which I denote [tex]R[/tex].
Parameterize the region by
[tex]T:\mathbf r(x,y,z(x,y))=\left\langle x,y,1+2x+3y+4y^2\right\rangle[/tex]
with [tex]1\le x\le8[/tex] and [tex]0\le y\le 1[/tex]. Then the area is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\left\|\frac{\partial\mathbf r}{\partial x}\times\frac{\partial\mathbf r}{\partial y}\right\|\,\mathrm dx\,\mathrm dy[/tex]
You have
[tex]\dfrac{\partial\mathbf r}{\partial x}=\left\langle1,0,2\right\rangle[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial y}=\left\langle0,1,3+8y\right\rangle[/tex]
and the cross product is
[tex]\langle1,0,2\rangle\times\langle0,1,3+8y\rangle=\langle-2,-3-8y,1\rangle[/tex]
with norm [tex]\sqrt{5+(3+8y)^2}[/tex].
So the integral reduces to
[tex]\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\sqrt{5+(3+8y)^2}\,\mathrm dx\,\mathrm dy[/tex]
The integrand is independent of [tex]x[/tex], so you have
[tex]\displaystyle7\int_0^1\sqrt{5+(3+8y)^2}\,\mathrm dy[/tex]
Let [tex]3+8y=\sqrt5\tan t[/tex], so that [tex]\mathrm dy=\dfrac{\sqrt5}8\sec^2t\,\mathrm dt[/tex]. Then the integral is equivalent to
[tex]\displaystyle\dfrac{7\sqrt5}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sqrt{5+(\sqrt5\tan t)^2}\sec^2t\,\mathrm dt[/tex]
[tex]\displaystyle\dfrac{35}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sec^3t\,\mathrm dt[/tex]
I'll leave the rest as an exercise for you, but you should end up with a result resembling
[tex]\dfrac{105}4\sqrt{\dfrac72}+\dfrac{35}{16}\ln\dfrac{9+2\sqrt{14}}5[/tex]
Parameterize the region by
[tex]T:\mathbf r(x,y,z(x,y))=\left\langle x,y,1+2x+3y+4y^2\right\rangle[/tex]
with [tex]1\le x\le8[/tex] and [tex]0\le y\le 1[/tex]. Then the area is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\left\|\frac{\partial\mathbf r}{\partial x}\times\frac{\partial\mathbf r}{\partial y}\right\|\,\mathrm dx\,\mathrm dy[/tex]
You have
[tex]\dfrac{\partial\mathbf r}{\partial x}=\left\langle1,0,2\right\rangle[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial y}=\left\langle0,1,3+8y\right\rangle[/tex]
and the cross product is
[tex]\langle1,0,2\rangle\times\langle0,1,3+8y\rangle=\langle-2,-3-8y,1\rangle[/tex]
with norm [tex]\sqrt{5+(3+8y)^2}[/tex].
So the integral reduces to
[tex]\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\sqrt{5+(3+8y)^2}\,\mathrm dx\,\mathrm dy[/tex]
The integrand is independent of [tex]x[/tex], so you have
[tex]\displaystyle7\int_0^1\sqrt{5+(3+8y)^2}\,\mathrm dy[/tex]
Let [tex]3+8y=\sqrt5\tan t[/tex], so that [tex]\mathrm dy=\dfrac{\sqrt5}8\sec^2t\,\mathrm dt[/tex]. Then the integral is equivalent to
[tex]\displaystyle\dfrac{7\sqrt5}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sqrt{5+(\sqrt5\tan t)^2}\sec^2t\,\mathrm dt[/tex]
[tex]\displaystyle\dfrac{35}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sec^3t\,\mathrm dt[/tex]
I'll leave the rest as an exercise for you, but you should end up with a result resembling
[tex]\dfrac{105}4\sqrt{\dfrac72}+\dfrac{35}{16}\ln\dfrac{9+2\sqrt{14}}5[/tex]
The exact area of the surface [tex]\rm z = 1+2x+3y+4y^2[/tex] is [tex]\rm \dfrac{105}{4}\sqrt{\dfrac{7}{2}}+\dfrac{35}{16} \;ln\dfrac{9+2\sqrt{14} }{5}[/tex] and this can be determined by integrating the given equation.
Given:
[tex]\rm z = 1+2x+3y+4y^2[/tex]
The exact area of the surface is given by the surface integral.
[tex]\rm \int\int_sdA=\int^1_0\int^8_1||\frac{\delta r}{\delta x}\times \frac{\delta r}{\delta y}||\;dx\;dy[/tex] --- (1)
where [tex]\delta r/\delta x[/tex] and [tex]\delta r/\delta y[/tex] is given by:
[tex]\dfrac{\delta r}{\delta x}=<1,0,2>[/tex]
[tex]\dfrac{\delta r}{\delta y}=<0,1,3+8y>[/tex]
The cross product is given by:
<1,0,2> [tex]\times[/tex] <0,1,3 + 8y> = <-2,-3 - 8y,1>
The integral given in expression (1) is reduced to:
[tex]\rm \int\int_sdA=\int^1_0\int^8_1\sqrt{5+(3+8y)^2} \;dx\;dy[/tex]
[tex]\rm = 7\int^1_0\sqrt{5+(3+8y)^2}\;dy[/tex]
Now, substitute the value of [(3+8y) = [tex]\sqrt{5}[/tex] tan t] and dy becomes:
[tex]\rm dy= \dfrac{\sqrt{3} }{8}sec^2t \;dt[/tex]
Now, the integral is reduced to:
[tex]\rm = \dfrac{7\sqrt{5} }{8}\int^{tan^{-1}\frac{11}{\sqrt{5} }}_{tan^{-1}\frac{3}{\sqrt{5} }}\sqrt{5+(\sqrt{5} tan\; t)^2}\;sec^2t\;dy[/tex]
Now, simplify the above integral in order to determine the exact area of the surface.
[tex]\rm \int\int_sdA=\dfrac{105}{4}\sqrt{\dfrac{7}{2}}+\dfrac{35}{16} \;ln\dfrac{9+2\sqrt{14} }{5}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/22008756
