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=3sqrt(x) , y=3 and 2y+3x=6.
It is relatively easy to find the intersections
y = 3 and 2y+3x = 6 intersect at (0,3)
y = (3/2)√x and y = 3 intersect at (4,3)
y = (3/2)√x and the line 2y+3x = 6 intersect at (1,3/2)
If you take vertical slices you must find the area in two parts, but if we take horizontal slices we can do it in one calculation , letting y run from 3/2 to 3
we have to solve 2y = 3√x for x ---> x = (4/9)y^2
and solve 2y+3x=6 for x ---> x = 2 - (2/3)y
width of region = (4/9)y^2 - 2 + (2/3)y
area = [integral] (4/9)y^2 - 2 + (2/3)y dy from 3/2 to 3
= [(4/27)y^3 - 2y + (1/3)y^2] from 3/2 to 3
= (4 - 6 + 3) - (1/3 - 3 + 3/4)
= 35/12
check my arithmetic
To sketch the region enclosed by the given curves, we need to start by identifying the points where the curves intersect. Let's find these points step by step:
1. The first curve is 2y = 3√(x). To simplify the equation, we can square both sides:
(2y)^2 = (3√(x))^2
4y^2 = 9x
2. The second curve is y = 3, which is a horizontal line parallel to the x-axis and intersects the y-axis at y = 3.
3. The third curve is 2y + 3x = 6. To simplify, we can solve for y:
2y = 6 - 3x
y = (6 - 3x)/2
Now, let's find the points of intersection:
1. Substitute y = 3 into the equation of the first curve:
4(3)^2 = 9x
36 = 9x
x = 4
Therefore, one point of intersection is (4, 3).
2. Substitute y = (6 - 3x)/2 into the equation of the third curve:
2(6 - 3x)/2 + 3x = 6
6 - 3x + 3x = 6
6 = 6
This equation is always true, which means the third curve is equivalent to a line with infinite points of intersection with the other curves.
Now let's sketch the region:
1. Plot the horizontal line y = 3.
2. Plot the curve 2y = 3√(x) (or 4y^2 = 9x after squaring). The point (4, 3) is on this curve.
Since the points of intersection are not clearly defined on the graph, we can approximate the region using the known intersection point and the shapes of the curves. The region appears to be a triangle-like shape with the point (4, 3) as one of its vertices.
To find the area of the region, we need to decide whether to integrate with respect to x or y. In this case, it is easier to integrate with respect to x.
To find the limits of integration, consider the x-values of the points of intersection. In this case, the region is bounded by the vertical lines x = 0 and x = 4.
So, we integrate with respect to x:
Area = ∫(from 0 to 4) [(3 - (6 - 3x)/2) - (3√(x)/2)] dx
Now you can find the integral and compute the area of the region.