Upload
alvin-pearson
View
219
Download
3
Tags:
Embed Size (px)
Citation preview
(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