Compute the surface area?

Question

Compute the surface area of the portion of the plane 3x+y+2z=6 inside the cylinder x^2+y^2=4. (Answer: 2pi*sqrt(14)) The formula for the surface area of a cylinder is A=2pi*r^2+2pi*r*h But how do I find the radius and height? This is a topic of surface integrals.

Answer 1

This was a little easier than I thought, but may or may not be useful since your chapter is to specifically use integration. 

The equation for the cylinder, x^2+y^2=4 is in the format a^2 + b^2 = r^2, therefore, r=2.  As for height, I'm not sure why you're asking for height.  To solve this, you can either do a double integration or a way that doesn't involve integration at all.  As far as the double integration, I only got to A(s)=SS_D sqrt(1+(-3/2)^2+(-1/2)^2) dA , -3/2 being f_x and -1/2 being f_y of the plane if you solve for z.  (SS is my poor shorthand for the two integrals).  But I don't actually have a cal3 book, so I didn't get very far, and you'd have to integrate the surface area.

So instead, I used the method without integration, in which I took the normal of the plane, n=(3,1,2) and the normal of the xy axis, (0,0,1), and found the angle between them. 

cos(a)=(3(0)+1(0)+1(2))/(sqrt(3^2+1^2+2^2)*sqrt(0^2+0^2+1^2))      from the cos(a)= n₁*n₂ / (|n₁||n₂|)

cos(a)=2/sqrt(14)

Since the cylinder is circular, we can use the area of a circle on the xy plane by projecting the area and set it up like this:

A_xy=Acos(a) 

A= A_xy / cos(a) = A_xy sqrt(14) / 2

Since the area of a circle is pi*r^2, we can fill that in for A_xy.  Remember, our r=2.

A= pi*r^2 *sqrt(14)/2 = 2*pi*sqrt(14)

Answer 2

If a surface is given by the equation F (x,y,z) = c, then the vector V = F_x i + F_{y ^j} + F_z k is normal to the surface.

                                                   ( F_x^2 i + F_y^2 j + F_z^2 k)^(1/2)

The equation is given by s = ∫ ∫ ------------------------------------------- dA

                                          R                          | F_z |

 

 

                    3x + y + 2z = 6

 

 

                  ( 9 + 1 + 4)^(1/2)

s = ∫∫----------------------------- dy dx

                           | 2 |

 

 

                                   2    (4-x^2)^(1/2)

= (1/2)(14)^(1/2)         ∫    ∫            dy dx

                                   0    0

 

Here a change of variable makes integration much easier.

x^2 + y^2 = 2^2 => x = 2 cos θ and y = 2 sin θ  and dy dx = r dr dθ

 

 

                          π/2       2

= (1/2)(14)^(1/2) ∫        ∫ r dr dθ                          

                           0        0

 

 

                                  π/2            2

= (1/2)(1/2)(14)^(1/2) ∫        r^2 |           dθ                                   

                                  0               0

 

 

                                      π/2

= (4)(1/2)(1/2)(14)^(1/2) ∫ dθ

                                      0

 

 

                                         π/2

= (4)(1/2)(1/2)(14)^(1/2) θ |

                                         0

 

= (4)(1/2)(1/2)(14)^(1/2) ( π/2)

= (14)^(1/2) ( π/2)

Answer 3

Your cylinder formula looks more like the formula for a circle, since there's no mention of a height, z.  You can get to the radius from the generic equation x²+y²=r²; the radius of the circle would appear to be 2. 

As for the height, I'm not sure if it matters.  If you assumed an infinite height of the cylinder, then the given plane cuts into it at some point (and possibly all the way across it).  The area of the plane inside the cylinder would be 4π (πr²) if it were parallel to the xy-plane on which the circle is defined, but since it is at an angle, the area inside the cylinder must be greater than that *if* it cuts all the way through the cylinder.