Asked by Jess
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.
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.
Answers
Answered by
Knights
Could you give us a diagram? OR a link to one or describe it?
Answered by
Jess
I spaced it out. This the picture.
h t t p : / / g o o. g l / A h g J X
h t t p : / / g o o. g l / A h g J X
Answered by
Steve
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
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
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