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.
3 answers
Could you give us a diagram? OR a link to one or describe it?
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
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
1 answer