I would like to know if my answers are correct:
Disclaimer: We are allowed to keep our answers in formula form
1. Use the washer method to find the volume of the solid that is generated by rotating the plane region bounded by y=x^2 and y = 2-x^2 about the axis y=-1
My Work: [pi((3-(1)^2)-((1)^2+1)^2] - [pi((3-(-1)^2)-((-1)^2+1)^2
My Answer: 0??
2. Use the washer method to find the volume of the solid that is generated by revolving the region bounded by the lines 3y = 2x, x=0, and y=2 about the y-axis
My Work: pi(9(2)^3/12)
My Answer: 72pi/12
3. Use the washer method to find the volume of the solid that is generated by rotating the region in the first quadrant bounded above by y = cosx and on the right x = pi/2 about the line y=2
My work: pi((4(pi/2) - ((2-(cos(pi/2))^3)/3)
My Answer: 11.36
4. Use the washer method to find the volume of the solid that is generated by revolving the region bounded by the curve y-x^3, y=0, and x=2 about the x-axis
My Work: pi(2^4/4)
My Answer = 4pi
I would greatly appreciate if you could let me know if my answer are correct. I am trying to practice as much as I can, and I would really like to know if I am understanding the material correctly. If the answer if wrong I would really appreciate if you could give me a clue as to where I can try to start to solve the problem. Thank You!!
#1 Zero? Really? That did not see strange to you? Too bad you didn't show your original integral.
The graphs intersect at (-1,1) and (1,1). Since the area of a disc is πr^2, the volume of a disc of thickness dx is πr^2 dx. So, your integral should be
v = ∫[-1,1] π(R^2-r^2) dx
where R = (2-x^2)+1 and r = x^2+1
Your expression indicates that you did π(R-r)^2
v = ∫[-1,1] π(((2-x^2)+1)^2-(x^2+1)^2) dx
= ∫[-1,1] π(8-8x^2) dx
= 8π∫[-1,1] (1-x^2) dx
= 8π(x - x^3/3) [-1,1]
= 8π[(1 - 1/3)-(-1+1/3)]
= 8π[4/3]
= 32π/3
Using symmetry, you could have simplified the evaluation a bit by using
v = 2∫[0,1] π(8-8x^2) dx
Just for grins, you can always consider checking your answer by using shells instead. The volume of a cylinder of radius r and height h and thickness dy is 2πrh dy.
The problem in this case is that if we take h to be the horizontal distance between the curves, the boundary changes at y=1. Both curves have two branches when solving for x.
y=2-x^2
x = ±√(2-y)
y=x^2
x = ±√y
So, when integrating from in the interval [0,1] we have
h = √y - (-√y) = 2√y
Again, we can use symmetry and let
h/2 = √y
Then on the interval [1,2]
h/2 = √(2-y)
So, now our integral becomes
v = 2∫[0,1] 2πrh dy + 2∫[1,2] 2πrh dy
where we change h at y=1. Luckily, r=y+1 in both parts. That leaves us with
v = 2∫[0,1] 2π(y+1)√y dy + 2∫[1,2] 2π(y+1)√(2-y) dy
= 4π∫[0,1] (y+1)√y dy + 4π∫[1,2] (y+1)√(2-y) dy)
= 4π(2/15 (3y+5) y^3/2)[0,1] + 4π(-2/5 (y+3) (2-y)^(3/2))[1,2]
= 64π/15 + 32π/15
= 96π/15
= 32π/5
#2
This time, our discs are of thickness dy, so
v = ∫[0,2] πr^2 dy
where r=x = 3y/2
v = π∫[0,2] (3y/2)^2 dy = 6π
checking with shells,
v = ∫[0,3] 2πrh dx
where r=x and h=2-y = 2 - 2x/3
v = ∫[0,3] 2πx(2 - 2x/3) dx
= 4π/3 ∫[0,3] 3x-x^2 dx
= 6π
Your answer, 72π/12 = 6π is correct.
#3
3. Use the washer method to find the volume of the solid that is generated by rotating the region in the first quadrant bounded above by y = cosx and on the right x = pi/2 about the line y=2
My work: pi((4(pi/2) - ((2-(cos(pi/2))^3)/3)
My Answer: 11.36
v = ∫[0,π/2] π(R^2-r^2)dx
where r=2-y=2-cosx and R=2
v = ∫[0,π/2] π(2^2-(2-cosx)^2)dx = π/4 (16-π) ≈ 10.099
using shells,
v = ∫[0,1] 2πrh dy
where r=2-y and h=x=arccos(y)
v = ∫[0,1] 2π(2-y)arccos(y) dy
= π/4 (16-π)
using discs of thickness dx,
v = ∫[0,2] πr^2 dx
where r=y=x^3
v = ∫[0,2] π(x^3)^2 dx
= ∫[0,2] πx^6 dx
= 128π/7
using shells of thickness dy,
v = ∫[0,8] 2πrh dy
where r=y and h=2-x = 2-∛y
v = ∫[0,8] 2πy(2-∛y) dy
= 128π/7
Just a note on #3
even though ∫x^2 dx = x^3/3
∫cos^2(x) dx is NOT cos^3(x)/3
cos^2(x) = 1/2 (1+cos(2x))
since cos(2x) = 2cos^2(x) - 1
Thus,
∫cos^2(x) dx = x/2 + 1/4 sin(2x)
Let's go through each problem to check your answers:
1. Use the washer method to find the volume of the solid that is generated by rotating the plane region bounded by y=x^2 and y = 2-x^2 about the axis y=-1.
To solve this, you need to integrate the difference of two volumes. The formula for the washer method is:
V = π∫[R(x)^2 - r(x)^2]dx
where R(x) is the outer radius and r(x) is the inner radius.
In this case, the outer radius is the distance from the axis of rotation (y = -1) to the curve y = 2 - x^2, which is 3. The inner radius is the distance from the axis of rotation to the curve y = x^2, which is 1.
To find the volume, you need to integrate the difference of two areas:
V = π∫[(3^2 - 1^2)dx]
= π∫[(9 - 1)dx]
= π∫[8dx]
= 8πx
Evaluating this integral from x = -1 to 1 gives:
V = 8π(1 - (-1))
= 8π(2)
= 16π
So, your answer for problem 1 should be 16π, not 0.
2. Use the washer method to find the volume of the solid that is generated by revolving the region bounded by the lines 3y = 2x, x=0, and y=2 about the y-axis.
To solve this problem, you also need to use the washer method and find the difference of two volumes.
The outer radius is the distance from the y-axis to the line 3y = 2x, which is x/2. The inner radius is the distance from the y-axis to the line y = 2, which is 2.
To find the volume, integrate the difference of two areas:
V = π∫[(x/2)^2 - 2^2]dx
= π∫[(x^2/4) - 4]dx
= π[(x^3/12) - 4x] | from x = 0 to x = 6
Evaluating this integral gives:
V = π[(6^3/12) - 4(6) - (0^3/12) + 4(0)]
= π[(6^3/12) - 24]
= π[72/4 - 24]
= π[18 - 24]
= -6π
So, your answer for problem 2 should be -6π, not 72π/12.
3. Use the washer method to find the volume of the solid that is generated by rotating the region in the first quadrant bounded above by y = cosx and on the right x = pi/2 about the line y=2.
To solve this problem, again, you'll need to find the difference of two volumes using the washer method.
The outer radius is the distance from the line y = 2 to the curve y = cosx, which is 2 - cos(x). The inner radius is the distance from the line y = 2 to the axis of rotation, which is 2 - 2 = 0.
To find the volume, integrate the difference of two areas:
V = π∫[(2 - cos(x))^2 - 0^2]dx
= π∫[(4 - 4cos(x) + cos^2(x))]dx
= π[(4x - 4sin(x) + (1/2)(x + sin(2x)))] | from x = 0 to x = π/2
Evaluating this integral gives:
V = π[(4(π/2) - 4sin(π/2) + (1/2)((π/2) + sin(π)) - 0)]
= π[2π - 4 + (1/2)(π/2) + 0]
= π[2π - 4 + π/4]
= (π/4)(8π + 1)
So, your answer for problem 3 should be (π/4)(8π + 1), not 11.36.
4. Use the washer method to find the volume of the solid that is generated by revolving the region bounded by the curve y-x^3, y=0, and x=2 about the x-axis.
To solve this problem, once again, you'll need to find the difference of two volumes using the washer method.
The outer radius is the distance from the x-axis to the curve y = x^3, which is x^3. The inner radius is the distance from the x-axis to the x-axis itself, which is 0.
To find the volume, integrate the difference of two areas:
V = π∫[(x^3)^2 - 0^2]dx
= π∫[(x^6)]dx
= (1/7)π(x^7) | from x = 0 to x = 2
Evaluating this integral gives:
V = (1/7)π(2^7 - 0^7)
= (1/7)π(128)
= (128/7)π
So, your answer for problem 4 is (128/7)π, not 4π.
In conclusion, your answers for problems 1, 2, 3, and 4 are incorrect. The correct answers are:
1. Volume = 16π.
2. Volume = -6π.
3. Volume = (π/4)(8π + 1).
4. Volume = (128/7)π.