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Chapter 13: Q 3. (page 1014)

Explain the difference between a type I region and a type II region.

Short Answer

Expert verified

It can be seen that, in the case of type I region, the integration is first done with respect to y and then with respect to x.

Step by step solution

01

Given Information

Two regions are given. We need to find the difference between two.

02

Consideration

In interval [a,b],a<b, type I region is bounded by y=g1(x)at the bottom and y=g2(x)at top for all x∈[a,b].

In interval[c,d],c<d, type I region is bounded by x=h1(y)to the left andx=h2(y)to the right for ally∈[c,d]

03

Simplification of first region 

The surface integral is calculated as ∬Ωf(x,y)dA

It can be calculated by taking an elemental area of breadth Δx, one end lying on y=g1(x)and other on y=g2(x)

Hence, surface integral is

∬Ωf(x,y)dA=∫abg1∫g1(x)g2(x)f(x,y)dydx

04

Simplification of second region 

It can be calculated by taking an elemental area of breadth Δyone end lying on x=h1(y)and other on role="math" localid="1653920549681" x=h2(y)

The integral becomes

∬Ωf(x,y)dA=∫ch4(y)dh2(y)f(x,y)dxdy

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Most popular questions from this chapter

Find the masses of the solids described in Exercises 53–56.

The solid bounded above by the plane with equation 2x + 3y − z = 2 and bounded below by the triangle with vertices (1, 0, 0), (4, 0, 0), and (0, 2, 0) if the density at each point is proportional to the distance of the point from the

xy-plane.

In Exercises 57–60, let R be the rectangular solid defined by

R = {(x, y, z) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2}.

Assume that the density at each point in Ris proportional to the distance of the point from the xy-plane.

(a) Without using calculus, explain why the x- and y-coordinates of the center of mass are x¯=2andy¯=32,respectively.

(b) Use an appropriate integral expression to find the z-coordinate of the center of mass.

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

∫∫Rxx+ydA,whereR={(x,y)|1≤x≤4and0≤y≤3}

Evaluate each of the integrals in exercise 33-36 as iterated integrals and then compare your answers with those you found in exercise 29-32

∫∫Rx2y3dAWhereR={(x,y)|1≤x≤3,0≤y≤2}

Evaluate the sums in Exercises 23–28.

∑i=14∑j=13∑k=12ij2k3
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