area - volume problems - crunchy...

Area - Volume Problems

1. The area of the region enclosed by the graph of y = X2 + 1 and the line y = 5 is

14a. -3

b.16-328c. -3

d. 323

e. 81r

2. What is the area of the region between the graphs of y = X2 and y = -x from x = 0 to x = 2 ?

2 ia.3

b. 8- .~3 ,.

c. 4

d. 14-316e.3

3. Let R be the region enclosed by the graph of y = 1+ ht( cos4x) the x-axis, and the lines x = -~ and x = ~. The

closest integer approximation of the area of R is

a. 0b. Ic. 2d. 3e. 4

4. What is the area of the region in the first quadrant enclosed by the graphs ofy =cos x, y =x, andthe y-axis?

a. 0.127b. 0.385c. 0.400d. 0.600e. 0.947

5. If 0 ~ k < ~ and the area under the curve y = cosx from x = k to x = ~ is 0.1, then k =

a. 1.471b. 1.414c. 1.277d. 1.120e. 0.436

._- ... --- ------

6. If the region enclosed by the y-axis, the line y =2, and the curvey = fx is revolved about they -axis, the volume of the solid generated is

32na. 5

b. 16n3

16nc. 5

d. 8n3

e. 'It

7. The base ofa solid S is the region enclosed by the graph ofy = ~, the line x = e. and thex-axis. If the cross sections of S perpendicular to the z-axis are squares, then the volume of S is

Ia. -2

b. 2- -j3c. I .;id. 2

e. ~ (e3 -I)

8. The base ofa solid is the region in the first quadrant enclosed by the graph ofy = 2 - X2 and thecoordinate axes. If every cross section of the solid perpendicular to the y-axis is a square, the volumeof the solid is given by .

f 2a. n (2-y) dy

b. f (2-y)dy

c. n{./2 (2 - X2 rdx

d. r./2 ( 2r2-x dx

e. r./2 (2-x2)dx

9. The region bounded by the graph of y = 2x _X2 and the x-axis is the base of a solid. For this solid, each crosssection perpendicular to the x-axis is an equilateral triangle. What is the volume of the solid?

a. 1.333b. 1.067c. 0.577d. 0.462e. 0.267

y

4

--~--~------~~.x810.

The base of a solid is a region in the first quadrant bounded by the x-axis, the y-axis, and the line x + 2y = 8, asshown in the figure above. If cross sections of the solid perpendicular to the x-axis are semicircles, what is thevolume of the solid?

a. 12.566b. 14.661c. 16.755 Id. 67.021e. 134.041

.;i]

11. The base of a solid is the region in the first quadrant bounded by the y-axis, the graph of y = tan -Ix, the horizontalline y =3, and the vertical line x = I. For this solid, each cross section perpendicular to the x-axis is a square. Whatis the volume of the solid?

a. 2.561b. 6.612c. 8.046d. 8.755e. 20.773

y

No~~zccs Ae, X:: «1.07

4. Let R be the region in the first quadrant enclosed by the graphs of y = 2x and y = x2, as shown in thefigure above.

(a) Find the area of R.

(b) The region R is the base of a solid. For this solid, at each x the cross section perpendiculsr to the x-axis

has area A(x) = sin(; x). Find the volume of the solid.

(c) Another solid has the same base R. For this solid, the cross sections perpendicular to the y-axis are squares.Write, but do not evaluate, an integral expression for the volume of the solid.

y

o-I

--~~--------~---------'r-~x

-2LAklA.l~+O~2-00'6 )<~ 4,~q1. Let R be the region bounded by the graphs of y = sin(Jrx) and y = x3 - 4x, as shown in the figure above.

-3


(b) The horizontal line y = -2 splits the region R into two parts. Write, but do not evaluate, an integralexpression for the area of the part of R that is below this horizontal line.

(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square.Find the volume of this solid.

(d) The region R models the surface of a small pond. At all points in R at a distance x from the y-axis, thedepth of the water is given by h(x) = 3 - x. Find the volume of water in the pond.

----- ----

y

--::+---r--+--- •..x

C,,-kv..lcd-o (L.

~b y:~ L\.5"Lf1. Let R be the shaded region bounded by the graph of y = In x and the line y = x - 2, as shown above.


(b) Find the volume of the solid generated when R is rotated about the horizontal line 'y'= -3.

(c) Write, bu.tdo not evaluate, an int~gral expression that can be used to find the volume ,~fthe solid generatedwhen R is rotated about the y-axis. ;~

.~,.

2009 API!)CALCULUS AB FREE-RESPONSE QUESTIONS (Form B)

No calculator.

y

1

2

--~--~~--~----~-----+--~xo 1 2 3 4

4. Let R be the region bounded by the graphs of y = .JX and y = I' as shown in the figure above.


(b) The region R is the base of a solid. For this solid, the cross sections perpendicular to the x-axis are squares.Find the volume of this solid.

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotatedabout the horizontal line y = 2.

;------ -- -------

Ap® CALCULUS AB2009 SCORING GUIDELINES

Question 4

Let R be the region in the first quadrant enclosed by the graphs of y = 2x and y

y = x2, as shown in the figure above.

(a) Find the area of R.(b) The region R is the base ofa solid. For this solid, at each x the cross

section perpendicular to the x-axis has area A( x) = sin ( ~ x). Find the

volume of the solid.(c) Another solid has the same base R. For this solid, the cross sections

perpendicular to the y-axis are squares. Write, but do not evaluate, anintegral expression for the volume of the solid.

4

3

2

2

(a) Area = J:(2X - x2) dx

= x2 _ .!.x31 x=2

3 x=Q

(b) Volume = L\in( ~ x) dx

2 (1r)1 x=Z= --cos -x1r 2 x=O

4=

(4 2(c) Volume = J

o(& - i) dy

{

I: integrand3: 1: antiderivative

1 : answer

{


1 : answer .

{2 : integrand3:1 : limits

© 2009 The College Board. All rights reserved.Visit the College Board on th84Web: www.collegeboard.com.


Question 1

y

--~--------~~--------~~-x-1

-2

-3

Let R be the region bounded by the graphs of y = sin(1l"x) and y = x3 - 4x, as shown in the figureabove.

(a) Find the area of R.(b) The horizontal line y = -2 splits the region R into two parts. Write, but do not evaluate, an integral

expression for the area of the part of R that is below this horizontal line.

(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is asquare. Find the volume of this solid.

(d) The region R models the surface of a small pond. At all points in R at a distance x from the y-axis,the depth of the water is given by hex) = 3 - x. Find the volume of water in the pond.

(a) sin(1l"x) = x3 - 4x at x = 0 and x = 2

Area = 5:(sin (1l"X) - (x3- 4x )) dx = 4

(b) x3 - 4x = -2 at r = 0.5391889 and s = 1.6751309

The area of the stated region is s: (-2 - (x3- 4x)) dx

f2 2

(c) Volume = o (sin(1l"x)-(x3 -4x)) dx=9.978

(d) Volume = 5:(3 - x) (sin (zr») - (x3- 4x)) dx = 8.369 or 8.370

{

I: limits3: 1: integrand

1 : answer

{I: limits .2'. 1: integrand

2 : { I : integrand1 : answer

{I: integrand2:I : answer

© 2008 The College Board. All rights reserved.Visit the College Board on the Web: www.collegeboard.com.


Let R be the shaded region bounded by the graph of y = In x and the liney = x - 2, as shown above.(a) Find the area of R.(b) Find the volume of the solid generated when R is rotated about the horizontal

line y = -3.

(c) Write, but do not evaluate, an integral expression that can be used to find thevolume of the solid generated when R is rotated about the y-axis.

Question 1

--+--+-~:....----x

In(x) = x - 2 when x = 0.15859 and 3.14619.Let S = 0.15859 and T = 3.14619

(a) Area of R= S;(ln(X)-(X-2))dx = 1.949

(b) Volume = 1r S; ((In (x) + 3)2 - (x - 2 + 3)2 ) dx

= 34.198 or 34.199

fT-2( 2)(c) Volume = 1r S-2 (y + 2)2 - (eY) dy

{

I: integrand3: 1: limits

1 : answer

{2 : integrand

3 :1 : limits, constant, and answer

3 : { 2 : integrand1 : limits and constant

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2

Ap® CALCULUS AB2009 SCORING GUIDELINES (Form B)

(a) Find the area of R.(b) The region R is the base of a solid. For this solid, the

cross sections perpendicular to the x-axis are squares.Find the volume of this solid.

(c) Write, but do not evaluate, an integral expression for the -!'----f-----,f-----!f-----!-_xvolume of the solid generated when R is rotated about 0the horizontal line y = 2.

Question 4

Let R be the region bounded by the graphs of y = -IX and Y

y = 1, as shown in the figure above.2

1 3 i 42

4 21X=4 4(a) Area = ( (-IX -~) dx = !:"x3/2 - ~ =

Jo 2 3 4 x=o 3

2 5/2 31X=4=~_~+~ 82 5 12 x=o 15

(c) Volume = "f((2 - ~)' - (2 - .IX)' ) dx

{


1 : answer

{


1 : answer

{I: limits and constant

3:2 : integrand

© 2009 The College Board. All rights reserved.Visit the College Board on theSNeb: www.collegeboard.com.

area - volume problems - crunchy...

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