wiki assignment - question 5
DESCRIPTION
This is my full solution to question 5 for the Wiki assignment in AP Calculus.TRANSCRIPT
Calculus Wiki AssignmentBy MrSiwWy
"Just Pebbles on the Beach" by FlickR user RobW
Question Number
"White 5 on Orange Dumpster by FlickR user YanivG
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = √x2y = x 21
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 2
2
4
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
2R = ∫(√x x )dx0
1
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
2R = ∫(√x x )dxR = ∫√x dx ∫x dx2
0
10
1
0
1
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
2R = ∫(√x x )dxR = ∫√x dx ∫x dxR = [2x /3] [x /3]3/2 3
20
10
1
0
1
0
1
0
1
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
2R = ∫(√x x )dxR = ∫√x dx ∫x dxR = [2x /3] [x /3]R = 2(1) /3 (1) /3
3/2 32
3/2 3
0
10
1
0
1
0
1
0
1
a)Firstly, using the two functions given, the points of intersection must be found in order to establish the integration interval.
y = x 21 y = √x2
y = y x = √x x = x x = 0, x = 1
1 224
The two functions intersect at x = 0, and x =1, therefore these are the lower and upper limits for the integral to find R.
Now, given the nature of both functions, the graph will look something like shown to the right.
y = x12
y = √x2
This shows that y is actually above y in the euclidean plane, which must be known in order to calculate R.
2 1
2R = ∫(√x x )dxR = ∫√x dx ∫x dxR = [2x /3] [x /3]R = 2(1) /3 (1) /3
3/2 32
3/2 3
0
10
1
0
1
0
1
0
1
R = 1/3
b)In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
This region will give the actual area of the solid, which will be the outer circle area (produced by spinning y ) minus the inner circle area (produced by spinning y ).
21
b)
The actual area of the crosssectional pieces of R can therefore be given by π[(y) (y) ].2
2 12
In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:This region will give the actual area of the solid, which will be the outer circle area (produced by spinning y2) minus the inner circle area (produced by spinning y1).
R
b)
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (shown in the previous slide).
In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
2 1
The region R, which is given by π[(y) (y) ].2 2
b) In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
The region R, which is given by π(y y).22 1
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (as shown before).
1V = π∫(√x (x ) )dx2 2
0
2
b) In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
The region R, which is given by π(y y).22 1
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (as shown before).
1V = π∫(√x (x ) )dxV = π∫(x x )dx
2 2
40
0
1
2
b) In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
The region R, which is given by π(y y).22 1
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (as shown before).
1V = π∫(√x (x ) )dxV = π∫(x x )dxV = π[x /2 x /5]
2 2
4
0
1
0
0
1
52
2
b) In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:
The region R, which is given by π(y y).22 1
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (as shown before).
1V = π∫(√x (x ) )dxV = π∫(x x )dxV = π[x /2 x /5]V = π[(1) /2 (1) /5]
2 2
4
0
1
0
0
1
522 5
2
b) In order to determine the volume of the solid generated when R is rotated about the xaxis, the washer method must be used. This is because each crosssectional piece will look like this:The region R, which is
given by π(y y).22 1
The solid generated by revolving R around the xaxis will look as depicted to the left. To determine it's volume, simply integrate from x = 0 to x = 1 to add up all the different crosssectional washers (as shown before).
V = π∫(√x (x ) )dxV = π∫(x x )dxV = π[x /2 x /5]V = π[(1) /2 (1) /5]
2 2
4
0
1
0
1
0
1
522 5
2
V = 3π/5
c)This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.
Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
2
2
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dx2 20
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dx
2 2
5/2 40
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dx
2 2
5/2 4
45/2
0
1
0
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dxV = π/4[x /2 4x /7 + x /5]
2 2
5/2 4
45/2
2 7/2 50
1
0
1
0
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dxV = π/4[x /2 4x /7 + x /5]V = π/4[(1) /2 4(1) /7 + (1) /5]
2 2
5/2 4
45/2
2 7/2 5
7/22 50
1
0
1
0
1
0
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dxV = π/4[x /2 4x /7 + x /5]V = π/4[(1) /2 4(1) /7 + (1) /5]V = π/4(35/70 40/70 + 14/70)
2 2
5/2 4
45/2
2 7/2 5
7/22 50
1
0
1
0
1
0
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dxV = π/4[x /2 4x /7 + x /5]V = π/4[(1) /2 4(1) /7 + (1) /5]V = π/4(35/70 40/70 + 14/70)V = π/4(9/70)
2 2
5/2 4
45/2
2 7/2 5
7/22 50
1
0
1
0
1
0
1
0
1
c) This is where the question delves into the realm of higher difficulty. The question itself might be hard to perceive at first but this is basically what it's saying.
Now R, which is enclosed by the two parabolas in the question, forms a solid which has crosssections that are circles just by itself (without rotation around an axis).
In order to determine the volume of R, all we must do is find the area of each crosssectional circle along the interval x = 0 and x = 1 multiplied by dx (given that the radius is the difference in ycoordinates between the two functions divided by two).
Since y = √x is on top, it gives the highest ycoordinate of the circle.Since y = x is the bottom function, it gives the lowest ycoordinate of the circle
Finding the difference between y and y yields the diameter of each circle. We just need to use the radius, so divide the difference by 2.
Since we can now find the area of each crosssectional circle, all we must do to find the volume is integrate along the interval R resides, as this will let dx represent the thickness of each circle and add up each of these cylinderlike shapes (dx * circle) along x = 0 through x = 1.
V = π∫[1/2(√x x )] dxV = π∫[1/4(x 2x + x )]dxV = π/4∫(x 2x + x )dxV = π/4[x /2 4x /7 + x /5]V = π/4[(1) /2 4(1) /7 + (1) /5]V = π/4(35/70 40/70 + 14/70)V = π/4(9/70)
2 2
5/2 4
45/2
2 7/2 57/22 5
V = 9π/280
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