(SEC. 7 .3 DAY ONE)

Volumes of Revolution

DISK METHOD

Example 1: Find the volume of a sphere with a radius of 2.

METHOD 1: GEOMETRY!!!!!!!!!!

𝑉=43

𝜋 𝑟3

𝑉=43

𝜋 (2)3

𝑉=323

𝜋

𝑉 ≈33.51𝑢𝑛𝑖𝑡𝑠3

Example 1: Find the volume of a sphere with a radius of 2.

METHOD 2: CALCULUS!!!!!!!!!!Step One: Write an equation to represent the edge of the shape (in this case: a CIRCLE!!!).

Step Two: Solve the equation for y.

What shape would result from rotating this “function” over the x-axis?

𝒙𝟐+𝒚𝟐=𝟒

A SPHERE!!!

𝒚=±√𝟒−𝒙𝟐

CIRCLE!

Step Three: Determine what shape cross-sections (made perpendicular to x-axis) of the sphere are.

𝒚=√𝟒−𝒙𝟐

𝒚=−√𝟒− 𝒙𝟐

Step Five: SUM up the area of all the possible cross-sections. In calculus, we SUM using an INTEGRAL!!!

Step Six: Evaluate the integral.

Compare our answer above to the one we got using the geometry formula!!WE GET THE SAME ANSWER!

CALCULUS WORKS!!!

Step Four: Write the equation for the area of one of the cross-sections (in terms of x).

𝒚=√𝟒−𝒙𝟐

𝒚=−√𝟒− 𝒙𝟐

𝐴=𝜋 𝑟2 𝐴=𝜋 (√ 4−𝑥2 )2 𝐴=𝜋 (4−𝑥2 )

Volume=

Volume=

=

=

=

≈33.51𝑢𝑛𝑖𝑡𝑠3

Example 2: Find the VOLUME of the solid formed by rotating the region bounded by the line around the x-axis on the interval [0,4].

What shape is this problem referring to?A CYLINDER!!!

Use GEOMETRY to find the volume of the cylinder.𝑉=𝜋 𝑟2h

Use CALCULUS to find the volume of the cylinder.𝑉=∫

0

4

𝜋 𝑟2𝑑𝑥

𝑉=∫0

4

𝜋 22𝑑𝑥

𝑉=4𝜋 𝑥|40

=

Example 3: Find the VOLUME of the solid formed by rotating the region bounded by the line around the x-axis on the interval

[0,1].

What shape is this problem referring to?

Use GEOMETRY to find the volume of the cylinder.

𝑉=13

𝜋𝑟2h

Use CALCULUS to find the volume of the cylinder.

𝑉=∫0

1

𝜋 𝑟2𝑑𝑥

𝑉=∫0

1

𝜋 (2 𝑥)2𝑑𝑥

𝑉=43

𝜋 𝑢𝑛𝑖𝑡𝑠3

A CONE!!

𝑉=13

𝜋 (2 )2(1)

Consider what shape one cross-section (taken perpendicular to the x-axis) of the solid would be. A CIRCLE!!

𝑉=𝜋 ( 43 𝑥3|10 )

So what happens the solid is not one we have a geometric formula

for?

2( )b

a

V f x dx ∫

You have to use CALCULUS!!! Here’s the basic formula:

Radius of circular cross-

section

Example 4: Find the VOLUME of the solid formed by rotating the region bounded by the line around the x-axis on the interval

[-1,1].

2( )b

a

V f x dx ∫

𝑉=𝜋∫−1

1

(𝑥3− 𝑥+1 )2𝑑𝑥

Evaluate this using your calculator…

𝑉 ≈6.76𝑢𝑛𝑖𝑡𝑠3

Example 6: Find the VOLUME of the solid formed by rotating the region bounded by the line around the x-axis on the interval .

𝑉=𝜋 ∫0

2𝜋

(2+sin 𝑥 )2𝑑𝑥

𝑉=9𝜋 2𝑢𝑛𝑖𝑡𝑠3

Example 5: Find the VOLUME of the solid formed by rotating the region bounded by the line , and around the y-axis .

¿ 𝜋∫− 3

3

❑

Evaluate this using your calculator…

Because the circular cross-sections will be horizontal, we will integrate this time with respect to y! This means the bounds for integration should be y-values and the function must be solved for x.

𝑉=𝜋∫𝑎

𝑏

( 𝑓 (𝑦 ))2𝑑𝑦 ( 16 𝑦+ 12 )

2

𝑑𝑦

Example 6: Find the VOLUME of the solid formed by rotating the region bounded by the line , and around the y-axis .

≈74.16𝑢𝑛𝑖𝑡𝑠3

Example 7: Find the VOLUME of the solid formed by rotating the region bounded by the line , and the x-axis around the x-axis .

The key here is that you HAVE to use TWO integrals!

We need to determine EXACTLY where the functions intersect first…√𝑥=6−𝑥𝑥=36−12𝑥+𝑥2

0=𝑥2−13𝑥+360=(𝑥−9)(𝑥−4)𝑥=4𝑎𝑛𝑑 9

1. TO ROTATE OVER A LINE OTHER THAN ONE OF THE AXES.

2. TO ROTATE AN AREA NOT FORMED BY ONE OF THE AXES.

We still need to learn….

HOMEWORK