Question

Find the area of the surface. the part of the plane 4x + 4y + z = 16 that lies inside the cylinder x2 + y2 = 9

Answer 1

To find the area of the surface that lies inside the given plane and cylinder, you need to determine the intersection curve between the two surfaces. This can be done by solving the equations simultaneously. Once you have the intersection curve, you can set up the double integral to find the area of the surface.

Step-by-step explanation:

To find the area of the surface that lies inside the given plane and cylinder, we need to determine the intersection curve between the two surfaces by solving the equations simultaneously.

Substituting the equation of the plane and cylinder, we get 4x + 4y + z = 16 and x^2 + y^2 = 9. Rearranging the equation of the plane, we have z = 16 - 4x - 4y. Substituting this into the equation of the cylinder, we get x^2 + y^2 = 9.

Now we have a system of two equations: x^2 + y^2 = 9 and z = 16 - 4x - 4y. We can solve these equations to find the intersection curve, which will give us the limits for integration to find the area. Once we have the limits, we can set up the double integral to find the area of the surface.

Answer 2

As a first step, convert to cylindrical coordinates. The equation
`x^2+y^2=9=3^2` is a clue to set


`x=3r\cos\theta`


`y=3r\sin\theta`

then have
`0\le r\le1` and
`0\le\theta\le2\pi`. Meanwhile, the plane equation tells you to use


`4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta`

So we parameterize the part of the plane within the cylinder with the vector-valued function


`\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)`

The surface integral giving the area of the surface is



`\displaystyle\iint_(\mathcal S)\mathrm dS`

where
`\mathcal S` denotes the surface in question, and the surface element
`\mathrm dS` is


`\mathrm dS=\left\|(\partial\mathbf s)/(\partial r)*(\partial\mathbf s)/(\partial\theta)\right\|\,\mathrm dr\,\mathrm d\theta`

We have


`(\partial\mathbf s)/(\partial r)=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))`


`(\partial\mathbf s)/(\partial\theta)=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))`


`\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9√(33)r\,\mathrm dr\,\mathrm d\theta`

So the area is


`\displaystyle\iint_(\mathcal S)\mathrm dS=9√(33)\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=1)r\,\mathrm dr\,\mathrm d\theta=18√(33)\,\pi`