Q42P Find the volume between the plan... [FREE SOLUTION]

Chapter 5: Q42P (page 248)

Find the volume between the planes z = 2x + 3y +6 and z = 2x + 7y + 8, , and over the square in the (x,y) plane with vertices (0,0) , (1,0) (0,1) (1,1) .

Short Answer

Expert verified

The volume obtained for the planes over the vertices is 4

Step by step solution

Definition of double integral and mass formula

The double integral of f(x,y) over the area A in the (x,y) plane as the limit of this sum, and write it as $鈭� {鈭�}_{A} f (x, y) dxdy$

Calculation of the volume integral bounded by the curve

Calculate the volume between the planes z = 2x + 3y + 6 and z = 2x + 7y + 8, and over the square in the (x,y) plane with vertices (0,0) , (1,0) (0,1) (1,1)

$l = {鈭�}_{0}^{1} dx {鈭�}_{0}^{1} dy {鈭�}_{2 x + 3 y + 6}^{2 x - 7 y + 8} dz = {鈭�}_{0}^{1} dx {鈭�}_{0}^{1} dy (4 y + 2) = {{鈭�}_{0}^{1} dx (2 y^{2} + 2 y)}_{0}^{1}$

Simplification and solving further.

${{鈭�}_{0}^{1} d x (2 y^{2} + 2 y)}_{0}^{1} = {鈭�}_{0}^{1} 4 d x = 4$

Therefore, the volume is 4.

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91影视

Find the volume between the planes z = 2x + 3y +6 and z = 2x + 7y + 8, , and over the square in the (x,y) plane with vertices (0,0) , (1,0) (0,1) (1,1) .

Short Answer

Step by step solution

Definition of double integral and mass formula

Calculation of the volume integral bounded by the curve

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