The function 4x - 2x^2
I did this problem myself by doing:
pi * integral from 0 to 2 of ( 4 - (4x - 2x^2)^2)dx
Is this method correct? I get 56 pi / 15 as a final answer. In one of my classes, the teaching assistant told me that's not how to do it. I didn't have enough time to explain that she's reading the question incorrectly, but I could be wrong. She suggested using the washer method, but why would I do that if there's only one function and I'm revolving it around y = 4? If I used washer with 4 and the function, it would be the area in between the two and rotated about the x-axis.
I rambled, is it correct?
I did this problem myself by doing:
pi * integral from 0 to 2 of ( 4 - (4x - 2x^2)^2)dx
Is this method correct? I get 56 pi / 15 as a final answer. In one of my classes, the teaching assistant told me that's not how to do it. I didn't have enough time to explain that she's reading the question incorrectly, but I could be wrong. She suggested using the washer method, but why would I do that if there's only one function and I'm revolving it around y = 4? If I used washer with 4 and the function, it would be the area in between the two and rotated about the x-axis.
I rambled, is it correct?
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A). If the region that is rotated is between the parabola and the x axis, then you will have to use washer method. Draw a vertical strip in the shaded region. Rotate that strip about the line y =4.
You should see a washer. INT [ pi(R^2)-pi(r^2)dx from 0 to 2.
R= the radius of the outer circle = 4
r= the radius of the smaller circle= 4-(4x-2x^2)=2x^2-4x+4
So (pi)INT [ 4^2- (2x^2-4x+4)^2 dx ] for (0,2)= 256pi/15
B). If the region that is rotated is between the parabola and the line y=4,
then R = 4-(4x-2x^2)=2x^2-4x+4
in this case you would integrate pi (R^2) by disk method.
(Pi) INT[ (2x^2-4x+4)^2 dx for (0,2)= 224pi/15
Notice that the second value (B) is the air space that is subtracted in A.
Hoping this helps!
You should see a washer. INT [ pi(R^2)-pi(r^2)dx from 0 to 2.
R= the radius of the outer circle = 4
r= the radius of the smaller circle= 4-(4x-2x^2)=2x^2-4x+4
So (pi)INT [ 4^2- (2x^2-4x+4)^2 dx ] for (0,2)= 256pi/15
B). If the region that is rotated is between the parabola and the line y=4,
then R = 4-(4x-2x^2)=2x^2-4x+4
in this case you would integrate pi (R^2) by disk method.
(Pi) INT[ (2x^2-4x+4)^2 dx for (0,2)= 224pi/15
Notice that the second value (B) is the air space that is subtracted in A.
Hoping this helps!