The area of a surface between a plane and a cylinder is evaluated using the integral [tex]\int\limits \int\limits^{}_D {\sqrt{(\frac{dz}{dx})^2 + (\frac{dz}{dy})^2 + 1} } \, dA[/tex]. So, the area of the given surface is [tex]\frac{4\pi}{3}\sqrt{14}[/tex]
Given that
[tex]x + 2y + 3z =1[/tex]
[tex]x^2 + y^2 = 4[/tex]
Make z the subject in [tex]x + 2y + 3z =1[/tex]
[tex]3z = 1 - x - 2y[/tex]
[tex]z = \frac{1}{3}(1 - x - 2y)[/tex]
The surface area is calculated using the formula:
[tex]Area = \int\limits \int\limits^{}_D {\sqrt{(\frac{dz}{dx})^2 + (\frac{dz}{dy})^2 + 1} } \, dA[/tex]
Where: [tex]dA = rdr \times d\theta[/tex]
[tex]z = \frac{1}{3}(1 - x - 2y)[/tex]
Calculate [tex]\frac{dz}{dx}[/tex]
[tex]\frac{dz}{dx}= \frac{1}{3}(0 - 1 - 2 \times 0)[/tex]
[tex]\frac{dz}{dx}= \frac{1}{3}(0 - 1 - 0)[/tex]
[tex]\frac{dz}{dx}= \frac{1}{3}(- 1)[/tex]
[tex]\frac{dz}{dx}= -\frac{1}{3}[/tex]
Calculate [tex]\frac{dz}{dy}[/tex]
[tex]\frac{dz}{dy}= \frac{1}{3}(0 - 0 - 2 \times 1)[/tex]
[tex]\frac{dz}{dy}= \frac{1}{3}(- 2)[/tex]
[tex]\frac{dz}{dy}= -\frac{2}{3}[/tex]
Because the plane is inside [tex]x^2 + y^2 = 4[/tex], then the region of z is:
[tex]D = \{(r,\theta) | 0 \le r \le \sqrt{4}, 0 \le \theta \le 2\theta\}[/tex]
[tex]D = \{(r,\theta) | 0 \le r \le 2, 0 \le \theta \le 2\theta\}[/tex]
[tex]Area = \int\limits \int\limits^{}_D {\sqrt{(\frac{dz}{dx})^2 + (\frac{dz}{dy})^2 + 1} } \, dA[/tex] becomes
[tex]Area = \int\limits^{2\pi}_0 \int\limits^{2}_0 {\sqrt{(\frac{-1}{3})^2 + (\frac{-2}{3})^2 + 1} } \, dA[/tex]
[tex]Area = \int\limits^{2\pi}_0 \int\limits^{2}_0 {\sqrt{\frac{1}{9} + \frac{4}{9} + 1} } \, dA[/tex]
Take LCM
[tex]Area = \int\limits^{2\pi}_0 \int\limits^{2}_0 {\sqrt{\frac{1+4+9}{9}} } \, dA[/tex]
[tex]Area = \int\limits^{2\pi}_0 \int\limits^{2}_0 {\sqrt{\frac{14}{9}} } \, dA[/tex]
Evaluate the square root of 9
[tex]Area = \int\limits^{2\pi}_0 \int\limits^{2}_0 {\frac{\sqrt{14}}{3} } \, dA[/tex]
Remove the constant
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 \int\limits^{2}_0 { dA}[/tex]
Recall that: [tex]dA = rdr \times d\theta[/tex]
So, we have:
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 \int\limits^{2}_0 { rdr \times d\theta}[/tex]
Integrate
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 { \frac{r^2}{2} |\limits^{2}_0 \ d\theta}[/tex]
Expand
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 { \frac{2^2 - 0^2}{2} \ d\theta}[/tex]
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 { \frac{4}{2} \ d\theta}[/tex]
[tex]Area = \frac{\sqrt{14}}{3}\int\limits^{2\pi}_0 { 2 \ d\theta}[/tex]
Integrate
[tex]Area = \frac{\sqrt{14}}{3}[2\pi]|\limits^{2\pi}_0[/tex]
Expand
[tex]Area = \frac{\sqrt{14}}{3}[2 \times 2\pi - 0][/tex]
[tex]Area = \frac{\sqrt{14}}{3}[4\pi - 0][/tex]
[tex]Area = \frac{\sqrt{14}}{3}[4\pi][/tex]
[tex]Area = \frac{4\pi}{3}\sqrt{14}[/tex]
Hence, the area of the surface is: [tex]\frac{4\pi}{3}\sqrt{14}[/tex]
Read more about the area of a surface between a plane and a cylinder at:
https://brainly.com/question/17193351
