Question

Please help me.Find the surface area of the plane x + 2y + 2z = 12 cut off by : (a) x = 0, y = 0, x = 1, y = 1

Answer 1

Answer:

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Step-by-step explanation:

First, let me know what do you think about given equations?

here, x+2y+2z=12 is a plane cutting X,Y&Z axes at 12,6,6 respectively.

x=0 is a plane i.e. YZ-plane & y=0 is XZ-plane.

x^2+y^2=16 is a circular cylinder with radius 4 and Z-axis as its axis.

Now trace all these and you will get some part of plane x+2y+2z=12 inside the cylinder b/w axes- planes. This is the portion for which to find the surface area and take projection of this portion on xy-plane , you will get 1/4 of circle of radius 4 with centre at origin.( say this circular region R)

Now, write the equation of plane in form of Monge's equation i. e. z=6-(x/2)-y and find it's partial derivatives with respect to x and y both.

formula for surface area is, S=∬ ds over R.

here, dS =(√p2+q2+1) dxdy;where p and q are partial derivatives of the Monge's equation with respect to x&y respectively.

p=-1/2 , q = -1 ; so ds=(3/2) dx dy

Now, S =3/2∬ dxdy over R

you can see, S= (3/2)area of 1/4 of circle of radius 4 with centre at origin.

therefore S =(1/4)24π = 6π