Let R be the region in the first quadrant bounded by the graph y=3-√x the horizontal line y=1, and the y-axis as shown in the figure to the right.
Please show all work.
1. Find the area of R
2. Write but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y=-1.
3. Region R is the base of a solid. For each y, where 1≤y≤3, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose height is half the length of its base. Write, but do not evaluate, an integral expression that gives the volume of the solid.
Could you give us a diagram? OR a link to one or describe it?
I spaced it out. This the picture.
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when y=1, x=4
so, the area is
a = ∫[0,4] (y-1) dx
= ∫[0,4] 2-√x dx
= 2x - 2/3 x^(3/2) [1,4]
= 8/3
using shells,
v = ∫[1,3] 2πrh dy
where r = 2+y and h = x = (3-y)^2
for the weird solid,
v = ∫[1,3] xh dy
where x = (3-y)^2 and h = x/2
To find the area of the region R in the first quadrant, we need to first determine the points of intersection between the curves y = 3-√x and y = 1.
Setting the two equations equal to each other, we have:
3 - √x = 1
Subtracting 1 from both sides, we get:
2 - √x = 0
Adding √x to both sides, we have:
√x = 2
Squaring both sides, we get:
x = 4
So the curves intersect at x = 4.
To find the area of R, we need to integrate the difference between the two curves from x = 0 to x = 4.
The area can be calculated using the following integral:
A = ∫(3 - √x - 1) dx
= ∫(2 - √x) dx
Integrating, we have:
A = [2x - 2/3x^(3/2)] from 0 to 4
Substituting the values, we get:
A = [(2*4) - 2/3(4)^(3/2)] - [(2*0) - 2/3(0)^(3/2)]
A = 8 - 16/3
A = 24/3 - 16/3
A = 8/3
So the area of R is 8/3 square units.
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To find the volume of the solid generated when region R is rotated about the horizontal line y = -1, we need to use the method of cylindrical shells.
The volume can be calculated using the following integral expression:
V = 2π ∫(radius) (height) (width) dy, where y varies from 1 to 3
For each y, the radius of the cylindrical shell is given by the equation y = 3 - √x.
To solve for x in terms of y, we square both sides:
y^2 = (3 - √x)^2
Expanding, we have:
y^2 = 9 - 6√x + x
Rearranging the terms, we get:
x = y^2 - 9 + 6√x
To find the height of the cylindrical shell, we know that it is half the length of its base. So the height is given by:
height = (y - 1)/2
The width of the cylindrical shell is infinitesimally small, represented by dy.
Therefore, the integral expression that gives the volume of the solid is:
V = 2π ∫(3 - √x)(y - 1)/2 dy, where y varies from 1 to 3.
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For each y, where 1 ≤ y ≤ 3, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose height is half the length of its base.
The base length is determined by the x-coordinate at each value of y, which can be obtained by solving the equation y = 3 - √x for x in terms of y.
So the length of the base is:
base = 3 - y
Therefore, the area of the rectangle is:
area = (base) * (height)
Substituting the values, we have:
area = (3 - y) * (1/2)
To calculate the volume of the solid, we need to sum up the areas of all the rectangles for each value of y in the range 1 to 3.
The integral expression that gives the volume of the solid is:
V = ∫(base) * (height) dy, where y varies from 1 to 3.