MATH1051 Lecture 8: Lecture 08
6.2 – Volumes
Volume
●How would you take the 3D volume of a 2D function?
○Rotate it around a line
●Example
○How would you take the volume of the area of 2 x,x xis and y xisy = − 2
1 − a−a
around the x-axis?
●What would it look like?
○The volume created by the area rotated around the x-axis looks like a cone
●How can we find the volume?
○You can use integrals to find the volume of the cone
●What is the area of a circle?
○rπ2
○What would the boundaries be?
■The x and y intercepts are where the function intercepts with the other
lines
■So 0, x 4x= =
○What is the radius?
■The function itself is the radius
■ (2 x)r= − 2
1
● πrΔxV = 2
○ (2 ) dxV = ∫
4
0
π − 2
12
○π 2x x dx = ∫
4
0
4 − + 4
1 2
○Here, you can move outside the integral and foil out π 2 )( − 2
12
○π(4x x x) (between 0 and 4) = − 2+ 1
12 3
■Then take the integral
○π(16 6 ) π(0 0 0) = − 1 + 12
64 − − +
■Plugging in the boundaries
○ = 3
16π
●Now you can check whether this is true using the actual formula for Volume of a cone
○ πr HV = 3
12
■Plug in the necessary components
○ (π)(2) (4)V= 3
12
○ V= 3
16π
Example # 1
Document Summary
How would you take the volume of the area of. The volume created by the area rotated around the x-axis looks like a cone. You can use integrals to find the volume of the cone. What is the area of a circle? r 2. The x and y intercepts are where the function intercepts with the other lines x = Here, you can move x ) (between 0 and 4) Now you can check whether this is true using the actual formula for volume of a cone. What would it look like? y = x and y = 2 x rotated around the x a xis. The unique thing about this function is that when rotated around the x-axis, there is an open space in the middle. Compile the volume after finding the boundaries and radii x = 2 x. Set equations equal to each other to find boundaries x x2 =
