Calculating Area of Enclosed Region

[vipertongn]

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?

 

[Mark44]

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:
\int_{-7}^6 f(x)dx + \int_{6}^{42} g(x)dx

An easier way is to use horizontal area elements. The area of a typical area element is [-y^2 + 42 - (-y)]\Delta y, and you need only one integral, and it's much easier to integrate.

 

[vipertongn]

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.

 

[Mark44]

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.

 

[vipertongn]

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...

 

[Mark44]

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 \Delta y. What is the interval along which the \Delta y's run?

Have you graphed the area yet?

 

[vipertongn]

I know it willl havea parabola at the top and a line underneath.