5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 301

12. Volume of solid by revolvingR:

VR=

∫ 4

0

π( 3 x)^2 dx=π

∫ 4

0

9 x^2 dx

=π

[

3 x^2

] 4

0 =^192 π

Set

∫ 4

0

π( 3 x)^2 dx=

192 π

2

⇒ 3 a^3 π= 96 π

a^3 = 32

a=( 32 )^1 /^3 = 2 ( 2 )^2 /^3

You can verify your result by evaluating

∫2(2) 2 / 3

0

π( 3 x)^2 dx.The result is 96π.

Part B Calculators are allowed.

13. (See Figure 12.9-10.)

y y = x 3

y = x^2

0 1 x

Figure 12.9-10

Step 1. Using the Washer Method:

Points of intersection: Set

x^3 =x^2 ⇒x^3 −x^2 = 0 ⇒

x^2 (x−1)=0orx=1.

Outer radius=x^2 ;

Inner radius=x^3.

Step 2. V=π

∫ 1

0

((

x^2

) 2

−

(

x^3

) 2 )

dx

=π

∫ 1

0

(x^4 −x^6 )dx

Step 3. Enter

∫

(π(x∧ 4 −x∧6),x,0,1)

and obtain

2 π

35

.

14. (See Figure 12.9-11.)

y

x

0

2

2

–2

–2

Figure 12.9-11

Step 1. x^2 +y^2 = 4 ⇒y^2 = 4 −x^2 ⇒

y=±

√

4 −x^2

Lets=a side of an equilateral

triangles= 2

√

4 −x^2.

Step 2. Area of a cross section:

A(x)=

s^2

√

3

4

=

(

2

√

4 −x^2

) (^2) √
3
4

.

Step 3. V=

∫ 2

− 2

(

2

√

4 −x^2

) 2 √ 3

4

dx

=

∫ 2

− 2

√

3(4−x∧2)dx

Step 4. Enter

∫ (√

(3)∗(4−x^2 ),x, −2, 2

)

and obtain

32

√

3

3

.