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The area between two curvesThe Volume of the Solid of revolution (by slicing)

1. AREA BETWEEN the CURVES

dA={( outerfunction)−( innerfunction)} dx

A=∫a

b

dA=∫a

b

[ y1 ( x )− y2(x )]dx A=∫c

d

dA=∫c

d

[ x1 ( y )−x2( y )] dy

EX: Determine the area of the region bounded by y = 2x2 +10 and y = 4x +16 between x = – 2 and x = 5

A=∫a

b

dA=∫a

b

[( outerfunction)−( innerfunction)]dx

APPLICATIONS OF THE DEFINITE INTEGRAL

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EX: Determine the area of the region enclosed by y = sin x and y = cos x and the y -axis for 0≤ x≤ π2

EX: Determine the area of the enclosed area by x=12y2−3 and y=x−1

Intersection: (-1,-2) and (5,4).

THE SAME: Determine the area of the enclosed area by x=12y2−3 and y=x−1

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So, in this last example we’ve seen a case where we could use either method to find the area. However, the second was definitely easier.

EX: Find area

Intersection points are:

y = - 1y = 3

A=∫−1

3

[ (− y2+10 )−( y−2 )2 ]dy

¿∫−1

3

[−2 y2+4 y+6 ] dy

¿ [−23y3+2 y2+6 y ]¿

A=643

Volume of REVOLUTION

▪ Find the Volume of revolution using the disk method▪ Find the volume of revolution using the washer method▪ Find the volume of revolution using the shell method▪ Find the volume of a solid with known cross sections

Area is only one of the applications of integration. We can add up representative volumes in the same way we add up representative rectangles. When we are measuring volumes of revolution, we can slice representative disks or washers.

DISK METHOD

V=∫a

b

dV dV=π r2dx

dV=π [ f (x ) ]2dx

dV=π [ f (x )+k ]2dx

dV=π [ f (x )−k ]2dx

dV=π [ k−f ( x ) ]2dx

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V=∫c

d

dV

WASHER METHOD

A solid obtained by revolving a region around a line.

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Volumes by Cylindrical Shells

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The Volume for Solids with Known Cross Sections

o Procedure: volume by slicing sketch the solid and a typical o cross section find a formula for the area, A(x), of the o cross section find limits of integration integrateo A(x) to get volume

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are

a) squares

b) Equilateral triangles

c) semicircles

x2 + y2 = 4 y=√4−x2

length of a side is : 2√4−x2

dV=A dx A=a2

V=4∫−2

2

( 4−x2 )dx=1283

x2 + y2 = 4 y=√4−x2

A=12a√a2−( a2 )

2

=√34a2=√3 ( 4−x2 )

V=∫−2

2

√3 ( 4−x2 )dx=32√3≈18.475

x2 + y2 = 4 y=√4−x2

A=12π ( a2 )

2

=18π a2=π 4−x2

2

d V=Adx

V=∫−2

2

π 4−x2

2dx=16 π

3≈16.755

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d) Isosceles right triangles

PRACTICE:

1. Find the volume of the solid generated by revolving about the x-axis the region bounded by the graph of f ( x )=√x−1 the x-axis, and the line x = 5. Draw a sketch. 1. ANS: 8π

2. Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph

of y=√cos x where 0≤ x≤ π2 the x-axis, and the y-axis. Draw a sketch. 2. ANS: π

3. Find the volume of the solid generated by revolving about the y-axis the region in the first quadrant bounded by the graph of y = x2, the y-axis, and the line y = 6. Draw a sketch. 3. ANS: 18 π

4. Using a calculator, find the volume of the solid generated by revolving about the line y = 8 the region bounded by the graph of y = x 2 + 4, the line y = 8. Draw a sketch. ANS: 512/15 π

5. Using a calculator, find the volume of the solid generated by revolving about the line y = –3 the region bounded by the graph of y = ex, the y-axis, the lines x = ln 2 and y = – 3. Sketch. 5. ANS: 13.7383 π

6. Using the Washer, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x in the first quadrant about the x-axis. Draw a sketch. Method (just a fancy name – use sketch

x2 + y2 = 4 y=√4−x2

A=12a

a2

tan π4

=a2

4=4−x2

d V=Adx

V=∫−2

2

(4−x2 )dx=323≈10.667

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and common sense!!! instead of given boundaries, you have to find it as intersection of two curves and then use sketch to subtract one volume from another ) 6. ANS: 4π/21

7. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x about the line y = 2. Draw a sketch. 7. ANS: 17π/21

8. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x2 and x = y2 about the y-axis. Draw a sketch. 8. ANS: 3π/10

AGAIN PRACTICE:

1. The base of a solid is the region enclosed by the ellipse x2

4+ y

2

25=1 .

The cross sections are perpendicular to the x-axis and are isosceles right triangles whose hypotenuses are on the ellipse. Find the volume of the solid. 1. ANS: V = 200/3

2. The base of a solid is the region enclosed by a triangle whose vertices are (0, 0), (4, 0) and (0, 2). The cross sections are semicircles perpendicular to the x-axis. Using a calculator, find the volume of the solid.

2. ANS: V = 2.094

3. Find the volume of the solid whose base is the region bounded by the lines x + 4 y = 4, x = 0, and y = 0, if the cross sections taken perpendicular to the x-axis are semicircles.

3. ANS: V = π/6

4. The base of a solid is the region in the first quadrant bounded by the y-axis, the graph of y = arctanx, the horizontal line y = 3, and the vertical line x = 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid? 4. ANS: V = ∫₀¹ (3 - arctan(x))² dx = 6.61233

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5. A solid has its base is the region bounded by the lines x + 2y = 6, x = 0 and y = 0 and the cross sections taken perpendicular to x-axis are circles. Find the volume the solid.

5. ANS: 9/2 π

6. A solid has its base is the region bounded by the lines x + y = 4, x = 0 and y = 0 and the cross section is perpendicular to the x-axis are equilateral triangles. Find its volume. 6. ANS: V = 16√3/3