Question

Find the area of the surface. the part of the cylinder y^2+z^2=9 that lies above the rectangle with vertices (0,0),(4,0),(0,2), and (4,2)

Answer 1

To find the area of the surface above the given rectangle, we need to determine the intersection of the cylinder with the plane of the rectangle. The surface area above the rectangle is equal to the area of the rectangle multiplied by the range of the z-coordinate.

Step-by-step explanation:

To find the area of the surface that lies above the given rectangle, we need to determine the intersection of the cylinder with the plane of the rectangle. The equation of the cylinder is y^2+z^2=9, and the vertices of the rectangle are (0,0), (4,0), (0,2), and (4,2). At the x=0 and x=4 cross sections, we can see that the y-coordinate ranges from 0 to 2 and the z-coordinate ranges from -3 to 3. Therefore, the surface area above the rectangle is equal to the area of the rectangle multiplied by the range of the z-coordinate.

The area of the rectangle is (4-0)(2-0) = 8 square units. The range of the z-coordinate is -3 to 3, so the surface area above the rectangle is 8 * (3 - (-3)) = 48 square units.

Answer 2

We can parameterize this part of the cylinder using


`\mathbf s(u,v)=\langleu,3\cos v,3\sin v\rangle`

with
`0\le u\le4` and
`\cos^(-1)\frac23\le v\le\frac\pi2`. Then the surface element is


`\mathrm d\mathbf S=\|\mathbf s_u*\mathbf s_v\|\,\mathrm du\,\mathrm dv`

`\mathrm d\mathbf S=3\,\mathrm du\,\mathrm dv`

so the area is given by the surface integral, where
`D` denotes the part of the cylinder in question,


`\displaystyle\iint_D\mathrm d\mathbf S=3\int_(v=\cos^(-1)(2/3))^(v=\pi/2)\int_(u=0)^(u=4)\mathrm du\,\mathrm dv=6\pi`

Probably less work would involve a simpler geometric approach, but it doesn't hurt to practice setting up a proper surface integral.