Question

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.

Answer 1

Could you give us a diagram? OR a link to one or describe it?

Answer 2

I spaced it out. This the picture.

h t t p : / / g o o. g l / A h g J X

Answer 3

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

Answer 4

[AI-generated answer — Explain Bot]

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.