Area Between Curves: Integrating Along The Y-Axis

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Hey guys! Today, we're diving into a classic calculus problem: finding the area of a region enclosed by two curves. But not just any curves – we're dealing with the equations F:x=y2+7F: x = y^2 + 7 and G:4y=19xG: 4y = 19 - x. The twist? We'll be partitioning the y-axis to solve this, which can sometimes be a more efficient approach. So, buckle up, grab your pencils, and let's get started!

Understanding the Curves

Before we jump into calculations, it's crucial to understand what these equations represent graphically.

  • F:x=y2+7F: x = y^2 + 7: This equation represents a parabola opening to the right. Notice that it's in the form x=ay2+cx = ay^2 + c, where 'a' is positive, indicating a rightward-opening parabola. The vertex of this parabola is at the point (7,0)(7, 0). Knowing the vertex is crucial because it gives us a starting point for sketching the graph. To get a better sense of the shape, we can also find a few other points. For instance, when y=1y = 1, x=8x = 8, and when y=1y = -1, x=8x = 8 as well. This symmetry is characteristic of parabolas. Let’s try y=2y=2, then x=22+7=11x = 2^2 + 7 = 11, and similarly when y=2y=-2, x=11x=11. These points help us visualize the parabola's width and how it curves. Understanding this parabola is crucial because it forms one of the boundaries of the area we're trying to calculate. We need a clear picture of its position and orientation to set up our integral correctly. Think of it like this: the parabola is the backbone of one side of our enclosed region, and we need to know its precise shape to measure the area accurately.
  • G:4y=19xG: 4y = 19 - x: This equation represents a straight line. We can rewrite it in the slope-intercept form (y=mx+by = mx + b) as y = - rac{1}{4}x + rac{19}{4}. This tells us that the line has a slope of - rac{1}{4} and a y-intercept of rac{19}{4}. To sketch this line, we can also find the x-intercept. Setting y=0y = 0, we get 0=19x0 = 19 - x, so x=19x = 19. Therefore, the line passes through the points (0, rac{19}{4}) and (19,0)(19, 0). Visualizing this line is just as important as understanding the parabola. The line acts as the other boundary of our enclosed region. Its slope and intercepts determine how it intersects the parabola, and these intersection points are what define the limits of integration for our area calculation. The interplay between the parabola and the line creates the shape whose area we want to find, so a good grasp of the line's position is essential.

By sketching these graphs (or using a graphing calculator), you'll see the region enclosed between the parabola and the line. This visual representation is super important for understanding the problem and setting up the integral correctly. Without a clear picture, it's easy to make mistakes in determining the limits of integration or which function is on top.

Finding the Intersection Points

Okay, we've got the curves visualized. Now, we need to find the points where they intersect. These points will define the limits of our integration along the y-axis. To find the intersection points, we need to solve the system of equations:

  1. x=y2+7x = y^2 + 7
  2. 4y=19x4y = 19 - x

We can substitute the expression for xx from equation (1) into equation (2):

4y=19(y2+7)4y = 19 - (y^2 + 7)

Now, let's simplify and rearrange this equation into a quadratic:

4y=19y274y = 19 - y^2 - 7

y2+4y12=0y^2 + 4y - 12 = 0

This quadratic equation can be factored nicely:

(y+6)(y2)=0(y + 6)(y - 2) = 0

So, the solutions for yy are:

y=6y = -6 and y=2y = 2

These are the y-coordinates of our intersection points. To find the corresponding x-coordinates, we can plug these y-values back into either equation (1) or (2). Let's use equation (1) as it's simpler:

  • For y=6y = -6: x=(6)2+7=36+7=43x = (-6)^2 + 7 = 36 + 7 = 43
  • For y=2y = 2: x=(2)2+7=4+7=11x = (2)^2 + 7 = 4 + 7 = 11

Therefore, the intersection points are (43,6)(43, -6) and (11,2)(11, 2). These points are super important because they define where the two curves meet. They essentially mark the boundaries of the region we're interested in. Imagine them as the anchors that hold our area in place. Without these points, we wouldn't know where to start and stop integrating, and our area calculation would be meaningless. So, finding these intersection points accurately is a critical step in solving the problem.

Setting Up the Integral

Here's where the magic happens! Since we're partitioning the y-axis, we'll be integrating with respect to yy. This means we need to express both equations in terms of yy, which we already have:

  • x=y2+7x = y^2 + 7 (Equation FF)
  • x=194yx = 19 - 4y (Rewritten Equation GG)

The area enclosed between the curves is given by the integral of the difference between the rightmost function and the leftmost function, with respect to yy, over the interval defined by the y-coordinates of the intersection points. In this case, the line (x=194yx = 19 - 4y) is to the right of the parabola (x=y2+7x = y^2 + 7) within the region of interest. So, our integral will look like this:

Area = egin{aligned} ewline \int_{-6}^{2} [(19 - 4y) - (y^2 + 7)] dy \end{aligned}

Let's break down why this integral is set up the way it is. First, the limits of integration, 6-6 and 22, come directly from the y-coordinates of the intersection points we calculated earlier. These values tell us the vertical extent of the region we're measuring. Next, the expression inside the integral, (194y)(y2+7)(19 - 4y) - (y^2 + 7), represents the horizontal distance between the two curves at a given value of yy. We're subtracting the parabola's x-value from the line's x-value because we want the width of the region at each yy. The larger this difference, the wider the region is at that particular y-value. This difference is what we're accumulating as we integrate. Think of it as summing up infinitely many tiny rectangles, each with a width of dydy and a length equal to the difference in x-values. The integral is simply the limit of this sum as the width of the rectangles approaches zero. Therefore, the correct setup of this integral is crucial for getting the right answer. It reflects our understanding of the geometry of the problem and how the area is formed between the curves.

Evaluating the Integral

Now, let's evaluate the integral to find the area. First, we simplify the integrand:

egin{aligned} ewline Area &= \int_{-6}^{2} (19 - 4y - y^2 - 7) dy \\ &= \int_{-6}^{2} (-y^2 - 4y + 12) dy \end{aligned}

Next, we find the antiderivative:

egin{aligned} ewline Area &= [- rac{1}{3}y^3 - 2y^2 + 12y]_{-6}^{2} \end{aligned}

Finally, we evaluate the antiderivative at the limits of integration:

egin{aligned} ewline Area &= [- rac{1}{3}(2)^3 - 2(2)^2 + 12(2)] - [- rac{1}{3}(-6)^3 - 2(-6)^2 + 12(-6)] \\\ &= [- rac{8}{3} - 8 + 24] - [- rac{1}{3}(-216) - 2(36) - 72] \\\ &= [- rac{8}{3} + 16] - [72 - 72 - 72] \\\ &= - rac{8}{3} + 16 + 72 \\\ &= 88 - rac{8}{3} \\\ &= rac{264 - 8}{3} \\\ &= rac{256}{3} \end{aligned}

Therefore, the area of the region enclosed by the graphs of the equations is rac{256}{3} square units. Calculating the integral is the final step in our journey, where we actually get the numerical answer for the area. Simplifying the integrand makes the antiderivative calculation easier and reduces the chances of errors. Remember, the antiderivative is the reverse process of differentiation, so we're looking for a function whose derivative would give us the expression inside the integral. Once we have the antiderivative, evaluating it at the limits of integration is a matter of plugging in the values and subtracting. This process gives us the definite integral, which represents the area we're seeking. The arithmetic might seem tedious, but accuracy is key here. A small mistake in the calculations can lead to a significantly different final answer. Therefore, carefully following each step and double-checking your work is essential to arrive at the correct area of rac{256}{3} square units.

Key Takeaways

  • Visualize the curves: Always start by sketching the graphs to understand the region you're working with.
  • Find the intersection points: These points define the limits of integration.
  • Set up the integral correctly: Pay attention to which function is on top (or to the right) and the limits of integration.
  • Evaluate carefully: Take your time and double-check your calculations.

This problem demonstrates a powerful application of calculus: finding the area between curves. By partitioning the y-axis, we were able to solve this problem efficiently. Keep practicing, and you'll become a pro at these types of problems in no time!

Practice Problems

To solidify your understanding, try these practice problems:

  1. Find the area enclosed by x=y2x = y^2 and x=2y2x = 2 - y^2.
  2. Determine the area of the region bounded by y=x2y = x^2 and y=xy = \sqrt{x}.

Remember, the key is to visualize, set up the integral correctly, and evaluate carefully. Good luck, and happy calculating!

Conclusion

So there you have it, guys! Finding the area enclosed between curves by partitioning the y-axis isn't as daunting as it might seem at first. By breaking down the problem into smaller, manageable steps – understanding the curves, finding the intersection points, setting up the integral, and evaluating it carefully – you can tackle these problems with confidence. Remember to always visualize the curves, as this is crucial for setting up the integral correctly. Practice makes perfect, so work through those practice problems and don't be afraid to ask for help when you need it. With a little bit of effort, you'll be mastering these calculus concepts in no time. Keep up the great work, and I'll catch you in the next one!