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LammettHash

See attached image for a plot of the region. Red represents the region of interest, which I denote \(R\).

Parameterize the region by

\(T:\mathbf r(x,y,z(x,y))=\left\langle x,y,1+2x+3y+4y^2\right\rangle\)

with \(1\le x\le8\) and \(0\le y\le 1\). Then the area is given by the surface integral

\(\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\left\|\frac{\partial\mathbf r}{\partial x}\times\frac{\partial\mathbf r}{\partial y}\right\|\,\mathrm dx\,\mathrm dy\)

You have

\(\dfrac{\partial\mathbf r}{\partial x}=\left\langle1,0,2\right\rangle\)
\(\dfrac{\partial\mathbf r}{\partial y}=\left\langle0,1,3+8y\right\rangle\)

and the cross product is

\(\langle1,0,2\rangle\times\langle0,1,3+8y\rangle=\langle-2,-3-8y,1\rangle\)

with norm \(\sqrt{5+(3+8y)^2}\).

So the integral reduces to

\(\displaystyle\iint_S\mathrm dA=\int_0^1\int_1^8\sqrt{5+(3+8y)^2}\,\mathrm dx\,\mathrm dy\)

The integrand is independent of \(x\), so you have

\(\displaystyle7\int_0^1\sqrt{5+(3+8y)^2}\,\mathrm dy\)

Let \(3+8y=\sqrt5\tan t\), so that \(\mathrm dy=\dfrac{\sqrt5}8\sec^2t\,\mathrm dt\). Then the integral is equivalent to

\(\displaystyle\dfrac{7\sqrt5}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sqrt{5+(\sqrt5\tan t)^2}\sec^2t\,\mathrm dt\)
\(\displaystyle\dfrac{35}8\int_{\arctan\frac3{\sqrt5}}^{\arctan\frac{11}{\sqrt5}}\sec^3t\,\mathrm dt\)

I'll leave the rest as an exercise for you, but you should end up with a result resembling

\(\dfrac{105}4\sqrt{\dfrac72}+\dfrac{35}{16}\ln\dfrac{9+2\sqrt{14}}5\)

View image LammettHash

The exact area of the surface \(\rm z = 1+2x+3y+4y^2\) is \(\rm \dfrac{105}{4}\sqrt{\dfrac{7}{2}}+\dfrac{35}{16} \;ln\dfrac{9+2\sqrt{14} }{5}\)  and this can be determined by integrating the given equation.

Given:

\(\rm z = 1+2x+3y+4y^2\)

The exact area of the surface is given by the surface integral.

\(\rm \int\int_sdA=\int^1_0\int^8_1||\frac{\delta r}{\delta x}\times \frac{\delta r}{\delta y}||\;dx\;dy\)    --- (1)

where \(\delta r/\delta x\) and \(\delta r/\delta y\) is given by:

\(\dfrac{\delta r}{\delta x}=<1,0,2>\)

\(\dfrac{\delta r}{\delta y}=<0,1,3+8y>\)

The cross product is given by:

<1,0,2> \(\times\) <0,1,3 + 8y> = <-2,-3 - 8y,1>

The integral given in expression (1) is reduced to:

\(\rm \int\int_sdA=\int^1_0\int^8_1\sqrt{5+(3+8y)^2} \;dx\;dy\)

           \(\rm = 7\int^1_0\sqrt{5+(3+8y)^2}\;dy\)

           

Now, substitute the value of [(3+8y) = \(\sqrt{5}\) tan t] and dy becomes:

\(\rm dy= \dfrac{\sqrt{3} }{8}sec^2t \;dt\)

Now, the integral is reduced to:

\(\rm = \dfrac{7\sqrt{5} }{8}\int^{tan^{-1}\frac{11}{\sqrt{5} }}_{tan^{-1}\frac{3}{\sqrt{5} }}\sqrt{5+(\sqrt{5} tan\; t)^2}\;sec^2t\;dy\)

Now, simplify the above integral in order to determine the exact area of the surface.

\(\rm \int\int_sdA=\dfrac{105}{4}\sqrt{\dfrac{7}{2}}+\dfrac{35}{16} \;ln\dfrac{9+2\sqrt{14} }{5}\)

For more information, refer to the link given below:

https://brainly.com/question/22008756

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