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Chapter 5: Q19P (page 247)

Above the square with vertices at, (0,0), (2,0),(0,2) and (2,2) and under the plane z = 8-x+y.

Short Answer

Expert verified

The required solution is 32.

Step by step solution

01

Definition of double integral

The double integral of f(x,y)over the areaA in the (x,y)plane as the limit of this sum, and we write it as鈭�鈭�Af(x,y)dxdy.

02

Calculation of the volume under the curve

The volume above the square with vertices at (0,0),(2,0),(0,2) and (2,2), and under the plane z=8-x+y.

First, integrate over z:

|=鈭�x=02鈭�y=02鈭�z=08-x+ydzdxdy=鈭�02dy鈭�02dx8-x+y

03

Further calculation of the volume of the bounded region

Now, integrate over x:

|=鈭�02dy8x-12x2+yx02=鈭�02dy(16-2+2y)

04

Further calculation of the volume of the bounded region

Now, integrate over y:

|=鈭�02dy(16-2+2y)=(14y+y2)02=32

Therefore, the value is 32

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Most popular questions from this chapter

In Problems 17 to 30, for the curve y=x, betweenx=0andx=2, find:

The moments of a thin shell whose shape is the curved surface of the solid (assuming constant density).

(a) Find the area of the surface z=1+x2+y2inside the cylinder x2+y2=1

(b) Find the volume inside the cylinder between the surface and the (x,y) plane. Use cylindrical coordinates

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer.

鈭�x-04鈭�y-0x/2y-dy-dx

鈭�Kdr1-K2r2=sin-1Kror-cos-1Krortan-1Kr1-K2r2

Hints: Sketch a right triangle with acute angles u and v and label the sides so that sinu=Kr. Also, note thatu+v=蟿蟿/2; then if u is an indefinite integral, so is -v since they differ by a constant of integration.

鈭�y-1/2dxdy over the area bounded by y=x2,x+y=2, and theaxis.
See all solutions

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