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Chapter 13: Q.52 (page 1028)

The region bounded above by the unit sphere centered at the origin and bounded below by the plane z=hwhere 0鈮�h鈮�1.

Short Answer

Expert verified

The solid's volume is bound.V=蟺h33

Step by step solution

01

Given information

The area is circumscribed on the top by the unit sphere centered at the origin, and on the bottom by the plane.z=h

02

simplification 

The goal of this issue is to create an iterated integral that depicts the volume of the region confined below the cone and above the plane z=husing polar coordinates.

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Most popular questions from this chapter

Evaluate each of the double integrals in Exercises 37鈥�54 as iterated integrals.

鈭�鈭�Rxx+ydA,whereR={(x,y)|1鈮�x鈮�4and0鈮�y鈮�3}

Evaluate the triple integrals over the specified rectangular solid region.

鈭�R鈥�x2yzdV, whereR={(x,y,z)鈭�鈭�2鈮�x鈮�1,0鈮�y鈮�3,and1鈮�z鈮�5}

Use Definition 13.4to evaluate the double integrals in Exercises 29鈥�32.

localid="1649936867482" 鈭�xx2ydA

whereR={(x,y)鈭�1鈮�x鈮�3and0鈮�y鈮�2}

Describe the three-dimensional region expressed in each iterated integral in Exercises 35鈥�44.

鈭�-15鈭�-32鈭�48f(x,y,z)dzdxdy

State Fubini's theorem.
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