(sec. 7.3 day one) volumes of revolution disk method
TRANSCRIPT
(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!!!!!!!!!!
π=43
π π3
π=43
π (2)3
π=323
π
π β33.51π’πππ‘π 3
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 βfunctionβ over the x-axis?
ππ+ππ=π
A SPHERE!!!
π=Β±βπβππ
CIRCLE!
Step Three: Determine what shape cross-sections (made perpendicular to x-axis) of the sphere are.
π=βπβππ
π=ββπβ ππ
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).
π=βπβππ
π=ββπβ ππ
π΄=π π2 π΄=π (β 4βπ₯2 )2 π΄=π (4βπ₯2 )
Volume=
Volume=
=
=
=
β33.51π’πππ‘π 3
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.π=π π2h
Use CALCULUS to find the volume of the cylinder.π=β«
0
4
π π2ππ₯
π=β«0
4
π 22ππ₯
π=4π π₯|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.
π=13
ππ2h
Use CALCULUS to find the volume of the cylinder.
π=β«0
1
π π2ππ₯
π=β«0
1
π (2 π₯)2ππ₯
π=43
π π’πππ‘π 3
A CONE!!
π=13
π (2 )2(1)
Consider what shape one cross-section (taken perpendicular to the x-axis) of the solid would be. A CIRCLE!!
π=π ( 43 π₯3|10 )
So what happens the solid is not one we have a geometric formula
for?
2( )b
a
V f x dx β«
You have to use CALCULUS!!! Hereβs 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 β«
π=πβ«β1
1
(π₯3β π₯+1 )2ππ₯
Evaluate this using your calculatorβ¦
π β6.76π’πππ‘π 3
Example 6: Find the VOLUME of the solid formed by rotating the region bounded by the line around the x-axis on the interval .
π=π β«0
2π
(2+sin π₯ )2ππ₯
π=9π 2π’πππ‘π 3
Example 5: Find the VOLUME of the solid formed by rotating the region bounded by the line , and around the y-axis .
ΒΏ πβ«β 3
3
β
Evaluate this using your calculatorβ¦
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.
π=πβ«π
π
( π (π¦ ))2ππ¦ ( 16 π¦+ 12 )
2
ππ¦
Example 6: Find the VOLUME of the solid formed by rotating the region bounded by the line , and around the y-axis .
β74.16π’πππ‘π 3
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 firstβ¦βπ₯=6βπ₯π₯=36β12π₯+π₯2
0=π₯2β13π₯+360=(π₯β9)(π₯β4)π₯=4πππ 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 learnβ¦.
HOMEWORK