Find the area of the part of the plane x +y + z = 6 that lies above the square with vertices (0, 0), (1, 0), (0, 1), and (1, 1).
A)1
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D)2
E) $\sqrt { 5 }$
F) $\sqrt { 6 }$
G) $\frac { \sqrt { 5 } } { 2 }$
H) $\frac { \sqrt { 6 } } { 2 }$

Question

Find the area of the part of the plane x +y + z = 6 that lies above the square with vertices (0, 0), (1, 0), (0, 1), and (1, 1).
A)1
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D)2
E) $\sqrt { 5 }$ 
F) $\sqrt { 6 }$ 
G) $\frac { \sqrt { 5 } } { 2 }$ 
H) $\frac { \sqrt { 6 } } { 2 }$

Quizplus · Accepted Answer

The area of the part of the plane x + y + z = 6 that lies above the square can be found by projecting this part onto the xy-plane and then calculating the area of the projection using the formula for the area of a parallelogram or by integrating. The given plane can be rewritten in terms of z as z = 6 - x - y, showing that for every point (x, y) in the square, there's a corresponding point (x, y, z) on the plane. The normal vector to the plane is (1, 1, 1), and the area of the projection is 1 (the area of the square). The actual area of the plane section is the area of the projection multiplied by the cosine of the angle between the normal to the plane and the z-axis. The cosine of this angle is the z-component of the unit normal vector, which is 1/√3 for the vector (1, 1, 1). Therefore, the area is 1 multiplied by √3, due to the correction factor for the angle between the plane and the vertical, resulting in √3.

Find the Area of the Part of the Plane X 2\sqrt { 2 }2​

Find the Area of the Part of the Plane X $\sqrt { 2 }$