Area of a region between two curves

[Casquibaldo]

Homework Statement

Set up sums of integrals that can be used to find the area of the region bounded by the graphs of the equations by integrating with respect to y.
y= sqrt(x) y=-x x=1 x=4

Homework Equations

∫[f(x) - g(x)] dx

The Attempt at a Solution

I did:
∫ (from 1 to 4) [sqrt(x) + x] dx and worked it out to get 73/6, but my friend said I'm doing it backwards and doesn't want to explain. If anyone could tell me what I'm doing wrong, I would be grateful.

 

[EWH]

Your answer and method appear to be correct.
[Edit: nevermind, I missed that it is integrating with respect to y.]

 

[Casquibaldo]

I would think so too, but he said something about it being respect to y, and I probably didn't take that class and it might mean that I have to switch something, I don't know.

 

[Mark44]

Homework Statement

Set up sums of integrals that can be used to find the area of the region bounded by the graphs of the equations by integrating with respect to y.[/color]
y= sqrt(x) y=-x x=1 x=4

Homework Equations

∫[f(x) - g(x)] dx

The Attempt at a Solution

I did:
∫ (from 1 to 4) [sqrt(x) + x] dx and worked it out to get 73/6, but my friend said I'm doing it backwards and doesn't want to explain. If anyone could tell me what I'm doing wrong, I would be grateful.

Your answer and method appear to be correct.

I haven't checked the answer, but the method is not correct, according to the problem statement. The OP has integrated with respect to x.

As Casquibaldo did it, the typical area element is a vertical strip of width Δx, with a height of √x + x.

To do this as the problem requires, the typical area element will be Δy, and you (Casquibaldo) will need a sum of three integrals, since the strips have different left and right endpoints.

You will also need to work with the inverses of the two functions, since at each endpoint you need the x-value as a function of y. To get you started, y = √x is equivalent to x = y².

 

[Casquibaldo]

oh ok so
x=-y
but how do I go about getting the points of intersection and from which side would I integrate it since it is all sideways ?

 

[EWH]

Tricky of them to specify the limits of integration as x=1 to x=4 when they want you to integrate with respect to y. (Assuming the problem is actually exactly as stated.) The area still has to be the same, though.

It's the same diagram first rotated 90 degrees counterclockwise, then reversed right-to left, and the axis labels interchanged. (Flip the paper over L-R and trace on the back side.) [Edit: feel free to ignore this paragraph, it's confusing.]

In the original equations, I'd split the integral horizontally into three bits:
from where y=-x intersects x=4 to where it intersects x=1;
from there to where x=1 intersects y=sqrt(x) (a rectangle);
and finally from there to where y=sqrt(x) intersects x=4.
There are also more complicated ways to split it up.

 

[Casquibaldo]

oh I see what you mean- thanks
And rest assured, I copied it just how it shows on my handout.

 

[Mark44]

Tricky of them to specify the limits of integration as x=1 to x=4 when they want you to integrate with respect to y. (Assuming the problem is actually exactly as stated.) The area still has to be the same, though.

It's the same diagram first rotated 90 degrees counterclockwise, then reversed right-to left, and the axis labels interchanged. (Flip the paper over L-R and trace on the back side.) [Edit: feel free to ignore this paragraph, it's confusing.]

This flipping of the paper is unnecessary if you understand functions and their inverses. If y = f(x) is a one-to-one function, then the equivalent equation x = f^-1(y) gives the inverse as a function of y. The two graphs are exactly the same; the only difference is that the first equation gives y in terms of x, while the second gives x in terms of y. Any point (x, y) on the first graph is also on the second.

A point of this problem seems to be to exercise one's familiarity with inverses.

In the original equations, I'd split the integral horizontally into three bits:
from where y=-x intersects x=4 to where it intersects x=1;
from there to where x=1 intersects y=sqrt(x) (a rectangle);
and finally from there to where y=sqrt(x) intersects x=4.
There are also more complicated ways to split it up.

 

[iRaid]

I'm not sure why anyone would want to do this with respect to y.. Just doesn't make sense to me. So if someone could post a logical reason, that'd be good.

 

[Mark44]

1. Quite often, integrating with respect to one variable is much easier than integrating with respect to another. What can be a difficult integral in one approach can be a piece of cake in another.
2. Testing your understanding of integration.
3. Testing your understanding of inverse functions.
4. Because the problem asked the OP to do it this way.

Are these enough?

 

[Casquibaldo]

Ok so now that I know what the integrand is
∫(y^2 +y)dy
how do I figure out from where to where it is? (like a and b)

 

[DryRun]

To find the limits easily, it's always a good idea to plot the graph. I've attached it to this post.

 

[Casquibaldo]

would that be from -4 to 2?
That means it would be -74/3 (which I think makes sense because most of the graph is negative

 

[Casquibaldo]

Oh wait is is rather
∫(sqrt(x))dx from 0 to 4 - ∫(-x)dx from 0 to 4
which is 40/3?
or ∫(sqrt(x))dx from 0 to 4 + ∫(-x)dx from 0 to 4
which is -8/3?

or is it just neither and I'm a bit confused

 

[DryRun]

In the original equations, I'd split the integral horizontally into three bits:
from where y=-x intersects x=4 to where it intersects x=1;
from there to where x=1 intersects y=sqrt(x) (a rectangle);
and finally from there to where y=sqrt(x) intersects x=4.

You need to specify different limits for each of the 3 integrals. Have you been able to identify the 3 split areas that will be added together to get the total area of the region?

Hint: start from the bottom of the graph and work your way up -- that's what the quoted description above says.
If you want to confirm that you have correctly recognized the 3 split areas, save the graph to your computer and color the 3 split areas and then attach it your reply.

 

[Casquibaldo]

I think this is what you mean

 

[Mark44]

Oh wait is is rather
∫(sqrt(x))dx from 0 to 4 - ∫(-x)dx from 0 to 4
which is 40/3?
or ∫(sqrt(x))dx from 0 to 4 + ∫(-x)dx from 0 to 4
which is -8/3?

or is it just neither and I'm a bit confused

No, both of these are wrong. You are still integrating with respect to x, and the integrals need to be done with respect to y. That means that the integrands need to be functions of y and dy needs to be in each integral.

I think this is what you mean

No, the regions you show are not the right regions.

The bottom region is triangular in shape, bounded by the line y = -1, the line y = -x, and the line x = 4.
The middle region is rectangular, bounded by the lines y = -1, y = 1, x = 1, and x = 4.
The top region is bounded by the lines y = 1 and x = 4 and the curve y = √x.

In each of the three regions, the typical area element has the form (x₂ - x₁)Δy. In each region you will need to use inverse functions to write x₁ and x₂ in terms of y.

 

[DryRun]

These are the correct regions that you need to find. I have shaded each of the 3 required regions a different color so you can easily distinguish between them.
You need to find the integral for each of these regions and then add them up.

 

[Mark44]

Yes, these are the regions I described in post #17.

 

[EWH]

The area will be the same as you calculated before. Here are the integration regions:

 

[DryRun]

The area will be the same as you calculated before. Here are the integration regions:

I've already said that in post #18.

 

[Casquibaldo]

this has become very clear to me, thank you :)
I think it would be something like
∫[-4, -1] (4-(-y))dy + ∫[-1, 1] (4-(1))dy + ∫[1, 2] (4-(y^2))dy

 

[Mark44]

Do you get the same area as before? Both integrations should yield the same result.

 

[Casquibaldo]

yes 73/6 :) thanks a lot everyone. The patience is definitely appreciated!

(it's the same as ∫[1, 4] (√x -(-x))dx )

 

[iRaid]

1. Quite often, integrating with respect to one variable is much easier than integrating with respect to another. What can be a difficult integral in one approach can be a piece of cake in another.
2. Testing your understanding of integration.
3. Testing your understanding of inverse functions.
4. Because the problem asked the OP to do it this way.

Are these enough?

I mean I know why integrating with respect to y is easier in some cases, but not with this problem..

 

[Mark44]

I mean I know why integrating with respect to y is easier in some cases, but not with this problem..

The problem isn't about which way is easier. As I said, I suspect that at least part of the region is to test the OP's skills at working with inverses and being able to look at this problem from a different perspective.