5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 277

Example 1

The base of a solid is the region enclosed by the ellipse
x^2
4

+

y^2

25

=1. The cross sections are

perpendicular to thex-axis and are isosceles right triangles whose hypotenuses are on the
ellipse. Find the volume of the solid. (See Figure 12.4-2.)
y

x

–2

2

0

5

–5

a

a

y

x^2 y^2

4 + 25 = 1

Figure 12.4-2

Step 1. Find the area of a cross sectionA(x).

Pythagorean Theorem:a^2 +a^2 =(2y)^2

2 a^2 = 4 y^2

a=

√

2 y, a> 0.

A(x)=

1

2

a^2 =

1

2

(√

2 y

) 2

=y^2

Since

x^2

4

+

y^2

25

=1,

y^2

25

= 1 −

x^2

4

ory^2 = 25 −

25 x^2

4

,

A(x)= 25 −

25 x^2

4

.

Step 2. Set up an integral.

V=

∫ 2

− 2

(

25 −

25 x^2

4

)

dx

Step 3. Evaluate the integral.

V=

∫ 2

− 2

(

25 −

25 x^2

4

)

dx = 25 x−

25

12

x^3

] 2

− 2

=

(

25(2)−

25

12

(2)^3

)

−

(

25(−2)−

25

12

(−2)^3

)

=

100

3

−

(

−

100

3

)

=

200

3

The volume of the solid is

200

3

.

Verify your result with a graphing calculator.