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=3(x^(1/2)) , y=4, and 2y+3x=6
I have been working on this problem for the past 3 hours with a friend and we have just hit a brick wall. We have literally done everything we possible can and cannot figure this out. Please help!!!
Thank you!
Draw the three curves
y1 = 4, y2 = -3/2 x +3 and y3 = (3/2)x^1/2
on the same graph to see what kind of a region you are dealing with.
The top border of the region is the horizontal y = 4 line extending from x = -2/3 to x = 64/9
There are two other bordering lines.
One is a straight line with negative slope extending downward from (-2/3, 4) to (1, 3/2). There is intersects the third line, which is the upper branch of a horizontal parabola extending from (1, 3/2) to (64/9, 4)
You can integrate either along x or y, but you would be integrating different functions. It looks a bit easier if you integrate it as the sum of two areas:
4 - (-3/2 x + 3) = 3/2 x +1 from x = -2/3 to 1, and
4 - (3/2)x^1/2 from x = 1 to x = 64/9
Always start a problem of this kind by sketching the region. This is something you will have to do yourself.
Thank you so much drwls!!! You do not understand how much I appreciate your help! You are a lifesaver! :) I not only got the answer, but also understand how to do it now.
To sketch the region enclosed by the given curves, we first need to graph each curve separately and then identify the region enclosed by them.
The given curves are:
1. 2y = 3√x
2. y = 4
3. 2y + 3x = 6
To decide whether to integrate with respect to x or y, we need to see how the curves are oriented. Let's start by graphing each curve separately.
1. 2y = 3√x:
To graph this equation, we can rewrite it as y = (3/2)√x. We can see that this curve is a square root graph, which starts from the point (0, 0) and increases as x increases.
2. y = 4:
This equation represents a horizontal line parallel to the x-axis at y = 4.
3. 2y + 3x = 6:
We can rearrange this equation to get y = (6 - 3x)/2, which represents a straight line with a slope of -3/2 and y-intercept of 3.
Now that we have graphed the three curves, let's determine the region enclosed by them. We are looking for the area of this region, so we need to find the limits of integration.
To do this, we need to find the x-values where these curves intersect. By substituting y = 4 into the other two equations, we can solve for x.
For the first equation:
2y = 3√x
2(4) = 3√x
8 = 3√x
(8/3)^2 = x
64/9 = x
For the third equation:
2y + 3x = 6
2(4) + 3x = 6
8 + 3x = 6
3x = -2
x = -2/3
So, the x-values where the curves intersect are x = -2/3 and x = 64/9.
Now that we have the limits of integration, we need to determine which variable to integrate with respect to. Since the region is bounded by vertical lines on the left and right sides, we will integrate with respect to x.
To find the area of the region, we integrate the difference between the top and bottom curve with respect to x over the given limits. The integral expression for the area is:
A = ∫[x₁, x₂] (top curve - bottom curve) dx
In this case, the bottom curve is y = 4 and the top curve is 2y = 3√x.
So, the area of the region is given by:
A = ∫[-2/3, 64/9] ([3√x/2] - 4) dx
Evaluating this integral will give you the area of the region enclosed by the given curves.