Find the area of the region.
2y=3sqrtx, y=5 and 2y+2x=5

ty

1. 👍 0
2. 👎 0
3. 👁 368
1. intersection points:
y = 5 with 2y = 3√x ---> (100/9 , 5)
y = 5 with 2x+2y = 5 ---> (-2.5, 5)
2y = 3√x with 2x + 2y = 5 ---> 1 , 3/2)

So I see it as
area
= ∫(5 - (5-2x)/2) dx from -2.5 to 1 + ∫((5-(3/2)(x^.5) )dx from 1 to 100/9

= 6.125 + 14.5185
= .....
not sure of the arithmetic again

1. 👍 0
2. 👎 0
2. Finishing Reiny's excellent evaluation, we get
49/8 + 392/27 = 4459/216 = 20.643

Now, if you want to use horizontal strips, there is no reason to divide up the region, since each strip is bounded on the left and right by a single curve.

Our two boundary curves are now

x = 4/9 y^2
x = (5-2y)/2
and we integrate over 3/2 <= y <= 5

∫[3/2,5] 4/9 y^2 - (5-2y)/2 dy
= 4/27 y^3 + 1/2 y^2 - 5/2 y [3/2,5]
= 4459/216 = 20.6435

http://www.wolframalpha.com/input/?i=plot+y+%3D+3%2F2+sqrt%28x%29%2C+2x%2B2y%3D5%2C+y%3D5

1. 👍 0
2. 👎 0

## Similar Questions

1. ### calculus

Sketch the region enclosed by the curves x=64−y^2 and x=y^2−64. Decide whether to integrate with respect to x or y. Then find the area of the region. Area =

2. ### calculus

Sketch the region in the first quadrant enclosed by y=6/x, y=2x, and y=12x. Decide whether to integrate with respect to x or y. Then find the area of the region.

3. ### Calculus AB

y=6-x y=x^2 Find the area of the region by integrating with respect to x. Find the area of the region by integrating with respect to y. ------------------------------------ i got the intersection pts to be(-3,9)and (2,4)....i then

4. ### math

Find the area of the region enclosed between y=2sin(x) and y=3cos(x) from x=0 to x=0.7π. Hint: Notice that this region consists of two parts.

1. ### Calculus

Sketch the region enclosed by the curves x= 49-y^2 and x = y^2 - 49. Decide whether to integrate with respect to x or y. Then find the area of the region.

2. ### calc

1. Let R be the region bounded by the x-axis, the graph of y=sqr(x) , and the line x=4 . a. Find the area of the region R. b. Find the value of h such that the vertical line x = h divides the region R into two regions of equal

3. ### Calculus-Area between curves

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days,

4. ### calculus

let R be the region bounded by the x-axis, the graph of y=sqrt(x+1), and the line x=3. Find the area of the region R

1. ### calculus

R is the first quadrant region enclosed by the x-axis, the curve y = 2x + a, and the line x = a, where a > 0. Find the value of a so that the area of the region R is 18 square units.

2. ### Calculus

The following are about an infinite region in the 1st quadrant between y=e^-x and the x-axis. A) Find the area of the region B)Find the volume of the solid generated by revolving the region about the y-axis

3. ### Calculus

Find the area of the region enclosed between y=4sin(x) and y=2cos(x) from x=0 to x=0.7pi. Hint: Notice that this region consists of two parts.

4. ### Calculus

The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3 A) Find the area of R B) B. Find the value of h, such that the vertical line x = h divides the region R into two Regions of equal area.