MA113_2010SPRING_HW2_PROFSOLN_[0]

MA113_2010SPRING_HW2_PROFSOLN_[0] - MA113 Calculus II The...

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Unformatted text preview: MA113 Calculus II The Cooper Union, Spring 2010 Prof. Alex Casti [email protected] Homework 2 Solutions Chapter 8 problems (Stein and Barcellos) Section 8.1 3 For the triangle bounded by y = 2 x , y = 3 x , and x = 1 , do the following: (a) Draw the region. See figure 1, which was generated with the Mathematica command RegionPlot . Figure 1: Region R associated with problem 3 a , section 8 . 1. (b) Compute the length of vertical cross sections c ( x ) . As is evident in figure 1, c ( x ) = 3 x- 2 x = x . (c) Compute the length of horizontal cross sections c ( y ) . The length c ( y ) of horizontal cross sections depends on whether the right-edge boundary is x = 1 or x = 1 2 y . If x = 1 then c ( y ) = 1- 1 3 y . If it is x = 1 2 y then c ( y ) = 1 2 y- 1 3 y = 1 6 y , and thus c ( y ) = 1 6 y , ≤ y ≤ 2 1- 1 3 y , 2 ≤ y ≤ 3 11 For the region bounded by y = x- 2 , y = , x = 1 , and x = 3 , sketch the region (see figure 2) and compute the following: Figure 2: Region R associated with problem 11, section 8 . 1. (a) Find the area by vertical cross sections. Each vertical cross section has length c ( x ) = x- 2 , so A = Z 3 1 x- 2 dx =- 1 x 3 x = 1 = 1- 1 3 = 2 3 . (b) Find the area by horizontal cross sections. The length c ( y ) of a horizontal cross section depends on whether the right-edge of a horizontal line ends at x = 3 or at x = 1 √ y . The cross-over point is y = 1 9 , so c ( y ) = 2 , ≤ y ≤ 1 9 y- 1 2- 1 , 1 9 ≤ y ≤ 1 → A = Z 1 9 2 dy + Z 1 1 9 y- 1 2- 1 dy = 2 9 + ( 2 √ y- y ) | 1 y = 1 9 = 2 9 + 2- 1- 2 3- 1 9 → A = 2 3 . You can see in this case that it was much simpler to com- pute the area with vertical cross sections. 1 17 Find the area between the curves y = sin x and y = cos x above the interval x ∈ , π 4 . Notice that cos x ≥ sin x over the given interval. It is easiest to use vertical cross sections with c ( x ) = cos x- sin x , and with this the area A between the curves is A = Z π 4 ( cos x- sin x ) dx = ( sin x + cos x ) | π 4 x = = √ 2 2 + √ 2 2-- 1 = √ 2- 1 . 27 Find the area A of the region bounded by y = tan x , y = , and x = π 4 (with x ≥ ) by vertical and horizontal cross sections. First, you should sketch tan x over the interval x ∈ , π 4 to see that the tangent function is positive over this domain. A vertical cross section will have length c ( x )= tan x , and a horizontal cross section will have length c ( y ) = π 4- tan- 1 y . Because tan π 4 = 1, the integration along the y-axis for the horizontal cross section calculation will run over the domain y ∈ [ , 1 ] . (a) Vertical cross sections A = Z π 4 tan xdx = ln | sec x | π 4 x = = ln √ 2- ln1 = ln2 1 2 → A = 1 2 ln2 . (b) Horizontal cross sections A = Z 1 π 4- tan- 1 y dy = π 4- y tan- 1 y- 1 2 ln ( 1 + y 2 ) 1 y = = π 4- tan- 1 1 + 1 2 ln2- + 1 2 ln1 = π 4- π 4 + 1 2 ln2 → A = 1 2 ln2 ....
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This note was uploaded on 02/27/2012 for the course CHEMISTRY/ CH/ECE/PH/ taught by Professor Faculty during the Spring '08 term at Cooper Union.

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MA113_2010SPRING_HW2_PROFSOLN_[0] - MA113 Calculus II The...

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