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 root(x), y=3, and 2y+2x=6

Answer 1

Q. 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√x, y=3, and 2y+2x=6.

Solution:

To find the area between the curves, we need to follow these steps:

1. Sketch the region

2. Find the points of intersection between the curves.

3. Determine the limits of integration.

4. Integrate the difference between the curves.

1.Sketch the region.

See sketch at the bottom.

2. Points of intersection:

    y = 2*√x and y = 3 intersect at x = (3/2)² = 9/4=2.25.

    y = 3 and y = 3 - x intersect at x = 0.

    y = 2*√x) and y = 3-x intersect at x = 1 and y=2.

          Steps:

1. 2*sqrt(x) = 3-x  
2.  (3-x)² =4x
3.  9-6x+x^2=4x
4. X^2-10x+9=0
5.  X=1 and x=9 (Not valid answer)

 

3.   Limits of integration:

   The x-limits are x = 0 and x = 3.

   The y-limits are y = 2 and y = 3.

4.  Integrate the difference:

 Now, integration can be done either wrt to x or y depending on your choice.

  The integration wrt y is simpler. 

Integration wrt y:

    y = 2*√x implies x = y²/4.

    y = 3-x implies x=3-y.

 

Hence;

Area =∫ (y²/4-(3-y)) dy between 2 and 3.

Area= 1.083

Integration wrt x:

Similarly;

Area = ∫ (3-(3-x)dx  between 0 and 1 + ∫ (3-(2*√x)dx  between 2.25 and 1

Area= 1.083

Note: Both answers are the same.

Answer 2

y= 4sqrx is the upward half of a right opening parabola with vertex at the orgin

y=3 is a flat horizontal line with y intercept (0,3)

2y+2x=6 or y=3-x is a downward straight line with intercepts (0,3) and (3,0)

set the y's equal to find points of intersection to use for limits of integration

3=4sqrx

3/4=sqrx

x = 9/16 (9/16,3)

3-x=3

x=0, y=3 (0,3)

3-x =4sqrx

16x = 9-6x+x^2

x^2-22x +9 = 0

x= 22/2+/-(1/2)sqr(22^2-36)

(.417, 2.582) ignore the other square root

then integrate using those points as limits to evaluate the integrals, depending on whether you integrate with respect to x or y, use the x or y coordinates of those points

with respect to xx

Area= about .096 which looks about in the right range if you graph the 3 lines, enclosed area is very small

integral of(3-4sqrx) dx evaluated between .417 and 9/16 =

plus

integral of (3-(3-x)dx = integral of x dx = x^2/2 evaluated from 0 to .417 = .087