Calculating Area of Enclosed Region

  • Thread starter vipertongn
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In summary: If you graph the antiderivative, you'll see that the interval corresponds to the y-values at the intersection points.
  • #1
vipertongn
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Homework Statement


x + y^2 = 42, x + y = 0

Find the area of the region enclosed.


Homework Equations



Integral of the top to the bottom.

The Attempt at a Solution



At first I tried to use S sqrt(42-x)+x dx and i get -2/3 (42 - x) ^ (3/2) - 1/2 x^2

However when I set them equal to one another 42-x=x^2 and get the root values -6 and 7. Plugging it in doesn't get me the right answer...what am I doing wrong?
 
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  • #2
Your first antiderivative looks right, but the second (and easier) one is off by a sign. It should be +1/2 x^2.

You should be working with limits of integration (a definite integral) if you want to come out with a number. To get the limits of integration you absolutely need a graph of the region. Have you done this?

You have two choices on how you can set up the integral: using vertical area elements or using vertical area elements. If you use vertical area elements, the top of each area element will always be y = +sqrt(42 -x), but the bottom of the elements are different depending on whether -7 <= x <= 6 or 6 <= x <= 42. In the first interval I listed, y = -x. In the second interval, y = -sqrt(42 -x). This means you need two definite integrals, set up like so:
[tex]\int_{-7}^6 f(x)dx + \int_{6}^{42} g(x)dx [/tex]

An easier way is to use horizontal area elements. The area of a typical area element is [itex][-y^2 + 42 - (-y)]\Delta y[/itex], and you need only one integral, and it's much easier to integrate.
 
  • #3
That makes sense, I can't get a picture of what it looks like though, however, what intervals would i use for the y then?

I set the x's equal to one another...
then I make it like -y^2+42+y and the values would be 6 and -7, but that isn't the case...I Plug them in and I end up with 119.6 and that's not the right answer.
 
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  • #4
You need a graph of the region to be able to get the limits of integration.
x + y^2 = 42 ==> y^2 = -x + 42 ==> y = +/-sqrt(-(x - 42))
You can graph this relation, can't you?

Concerning the work you show, you have
x = -y^2 + 42
x = -y
==> -y^2 -y + 42 = 0 ==> y^2 + y - 42 = 0
Solving this quadratic gives you y = -7 or y = 6.
These are the y values at the points of intersections of the two graphs. You can use either equation to find the corresponding x values.

In your first post you attempted to find an antiderivative of something, and then you apparently substituted y = -7 and y = 6 into the antiderivative. That makes no sense, which is why you didn't get the right answer.
 
  • #5
So I should find the x values to get the intervals used in the FTC for -y^2+42+y. i recall its integral right minus leftt and isn't that how I should do it. I'm not sure what I am doing wrong still...
 
  • #6
No, if you go with the easier tactic I described in post #2, you need the y-values at the two intersection points.

For my typical area element, the length (horizontal) is from the x value on the line to the x value on the parabola. The width (vertical) is [itex]\Delta y[/itex]. What is the interval along which the [itex]\Delta y[/itex]'s run?

Have you graphed the area yet?
 
  • #7
I know it willl havea parabola at the top and a line underneath.
 

1. How do I find the area of a rectangular region?

To find the area of a rectangular region, you simply need to multiply the length by the width. This will give you the total area in square units.

2. What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is 1/2 x base x height. Make sure to use the same unit of measurement for both the base and height when plugging them into the formula.

3. How do I find the area of a circle?

To find the area of a circle, you can use the formula A = πr², where A is the area and r is the radius of the circle. Alternatively, you can also use the formula A = πd²/4, where d is the diameter of the circle.

4. Can I use the same formula to find the area of any shape?

No, the formula for finding the area will vary depending on the shape of the region. For example, the formula for finding the area of a trapezoid is 1/2 x (base 1 + base 2) x height. It is important to use the correct formula for each specific shape.

5. How do I find the area of a region with irregular shapes?

In order to find the area of a region with irregular shapes, you can break it down into smaller, regular shapes such as rectangles and triangles. Find the area of each smaller shape and then add them together to get the total area of the irregular shape.

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