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Prove that the given surface area of the plane is cut off by the give planes?
totally lost on what to do here, if some one could explain how to do it with steps that would be great
Question
Prove that the surface area of the plane x + 2y + 2z = 12 cut off by the planes x=0,y=0,x=1,and y=1 is 3/2
so in other words evaluate the double integral with x bounds [0,1] and y bounds [0,1]
1 Answer
- Captain MephistoLv 710 years agoFavorite Answer
Vectors. Set up two vectors and then take the cross product. The magnitude of the cross product is the area. Use the given equation and the bounding planes to get the endpoints of the segments bounding the area and then use these to get the vectors.
V1 = side 1 = (1, 0, -1/2) .... in y = 0 plane, points are (0,0,6) and (1,0,5.5)
V2 = side 2 = (0, 1, -1) .... in x = 0 plane, points are (0,0,6) and (0,1,5)
V1xV2 = (1/2, 1, 1)
||V1xV2|| = sqrt(1/4 + 1 + 1) = sqrt(9/4) = 3/2
If this isn't clear enough, let me know and I will go into more details