# Integrals with Respect to X and Y

• silicon_hobo
In summary: Suppose I transposed the functions so that the whole area was within a positive domain and range. Then my method would work, right? Of course that would be more trouble than just using the absolute values.Since the line y=2x+4 crosses the x-axis at the point -2, whenever we need to find an area that this line encloses with the x-axis there will always be one limit of integration -2. So, on this interval, y is greater than or equal to 0 and so we would use the limit of integration -2.
silicon_hobo
[SOLVED] Integrals with Respect to X and Y

Hey folks, I've been working on this problem for too long.

## Homework Statement

Sketch the area of the region bounded by the curves $$y^2+4x=0$$ and $$y=2x+4$$. Set up two integrals, one with respect to x and one with respect to y, for finding the area of the region. Evaluate one of the integrals to find the area.

## Homework Equations

http://www.mcp-server.com/~lush/shillmud/graph1.5.JPG

I've got it drawn but I just can't get the integrals to match. The question says to only do one but I want to check my answer (and hopefully learn how to set these up). Here are my two latest attempts.

## The Attempt at a Solution

http://www.mcp-server.com/~lush/shillmud/inta1.5.JPG

http://www.mcp-server.com/~lush/shillmud/intb1.5.JPG

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One obvious point: the points of intersection have y= -4 and -1 but when y= -1, x is NOT 2!

Thanks. That fixed the numerator, bringing me somewhat closer, but now the denominators are not matching. I have no idea what's going on here.

You once again set the limits of integration incorrectly, look what happens if you proceede how u started.

$$\int_{-4}^{-1}(2x+4)dx=(x^{2}+4x)|_{-4}^{-1}=(1-4)-(16-16)=-3$$

which doesn't actually make sense, since the area cannot be negative. Instead
$$\int_{-4}^{-2}|(2x+4)|dx=-\int_{-4}^{-2}(2x+4)dx=-(x^{2}+4x)|_{-4}^{-2}= -[(4-8)-(16-16)]=5$$

YOu have to be carefule, because as i explained in another post, the $$\int_a^bf(x)dx$$ only calculates the area that the curve of the function f(x) encloses with the x-axis, meaning that you have to set up a different integral with the limits of integration wherever the graph crosses the x-axis.

Now which part of the region enclosed by those functions do you want to find by setting an integral with respect to y? In other words do you want to find the area that the two graphs enclose above x-axis or below x-axis, or do you want to do the whole thing with respect to y?

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The total enclosed area above and below the axis.

silicon_hobo said:
The total enclosed area above and below the axis.

I think you misunderstood me. I said that which part of that area would you like to find integrating with respect to y?
Or did i misunderstand you?

The whole thing wrt x and the whole thing wrt y.Thanks again.

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silicon_hobo said:
I want to find the total area between the line and curve by integrating wrt x & wrt y, all of it. Why did you switch the right boundry of the domain to -2 instead of -1 in your last example there? Thanks again.
Since the line $$y=2x+4$$ crosses the x-axis at the point -2, whenever we need to find an area that this line encloses with the x-axis there will always be one limit of integration -2. SInce like i said also in the previous post :
YOu have to be carefule, because the $$\int_a^bf(x)dx$$
only calculates the area that the curve of the function f(x) encloses with the x-axis, meaning that you have to set up a different integral with the limits of integration wherever the graph crosses the x-axis.

So say we want to calculate the area that the line f(x)=2x+4 encloses with the x-axis from -8 to 4 then we have:

$$A=\int_{-8}^{-2}|f(x)|dx+\int_{-2}^{4}f(x)dx$$ We have to proceede this way, among others, because the function f(x) from -8 to - is negative, notice it lies under the x-axis, so since we want the total area that f(x) encloses with the x-axis, we need to add the area that f(x) encloses from -8 to -2 and the area that f(x) encloses from -2 to 4. If we would not separate the integral this way, we would not end up adding these two areas but instead subtracting them, since like i mentioned f(x) from -8 to -4 has a neg. sign.

Hope this helps!

Suppose I transposed the functions so that the whole area was within a positive domain and range. Then my method would work, right? Of course that would be more trouble than just using the absolute values.

If I understand what you're saying then these previous problems should also be wrong because the bounded area crosses the x-axis and I have not used absolute values for the piece underneath. However, I was pretty sure they were correct, especially the triangle which I even checked with Heron's formula... hence my confusion.

http://www.mcp-server.com/~lush/shillmud/quest3.jpg

http://www.mcp-server.com/~lush/shillmud/quest4.jpg

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silicon_hobo said:
Suppose I transposed I transposed the functions so that the whole area was within a positive domain and range. Then my method would work, right? Of course that would be more trouble than just using the absolute values.
It does not have to be within a positive domain, only within a positive range. Let's reffere once more to the line you have y=2x+4, and say we want to find the area that this line encloses with the x-axis from -2 to 7. On this whole interval y is greater than or equal to 0( it is equal to zero only when x=-2). This way we would have:

$$A=\int_{-2}^7(2x+4)dx$$

silicon_hobo said:
Suppose I transposed the functions so that the whole area was within a positive domain and range. Then my method would work, right? Of course that would be more trouble than just using the absolute values.

With transposing a function, do you mean like for ex, shifting it upwards, downwards and other such stuff? If so, then you cannot apply the exact procedure, since the transposing function is itself a function, so we have to treat it as a function also, unless it is symetric say around y-axis, and we only reflect it around it.

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Bump. So does that mean the questions I edited in above are also wrong?

silicon_hobo said:
Bump. So does that mean the questions I edited in above are also wrong?
Let me check it, i did not notice that you edited your post. I'll see what i can do!

well, i am wondering how come you got the right answer on the first one, because you did not set the first integral correctly. But the answer is correct, i checked it also integrating with respect to y. But i will see later, what happened there, because i need to run now.

and for the second one, it looks fine to me. Moreover, you have taken in consideration the minus sign, but maybe you are just not aware of it, look at it closely. YOu could have also done the last one more quickly, and easier.

Thank you again for taking the time to explain this. I will go back and see if I can get it worked out.

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After a few tries I was able to solve wrt using substitution. Thanks again.

silicon_hobo said:
http://www.mcp-server.com/~lush/shillmud/quest4.jpg
Like i said before you have taken into consideration the minus sign here. When u set up the first integral you did it like this,

$$A=\int_{-2}^0[(x+2)-(x^{2}-4)]dx=\int_{-2}^0(x+2)dx-\int_{-2}^0(x^{2}-4)dx$$
Here at the very first sight it looks like you are subtracting the second integral from the first one, but indeed you are adding the area that the second integral calculates with the area that the first integral calculates, for the fact that the function $$f(x)=x^{2}-4$$ is negative from -2 to 0. So, this is actually what i meant when i said u have taken into consideration the minus sign. Because here you should be adding areas, instead of subtractig them. You have done the same thing with the second integral, which is correct.

Yep, it all makes sense now. Thanks for the followup... and now on to the next concept I don't understand

## What is the definition of an integral with respect to x and y?

An integral with respect to x and y is a mathematical concept that represents the accumulation of a specific quantity over a certain region in the x-y plane. It involves finding the area under a curve in the x-y plane and can be used to solve various problems in calculus and other areas of mathematics.

## How is an integral with respect to x and y different from a regular integral?

An integral with respect to x and y is a double integral, meaning it involves integrating with respect to two variables (x and y) instead of just one. This allows for the analysis of functions that vary in two dimensions, whereas a regular integral only looks at functions that vary in one dimension.

## What is the process for solving an integral with respect to x and y?

The process for solving an integral with respect to x and y involves first setting up the integral by determining the limits of integration and the integrand (the function being integrated). Then, the integral is evaluated using various techniques such as substitution, integration by parts, or partial fractions. Finally, the result is simplified to obtain a numerical answer or a general solution.

## What are some applications of integrals with respect to x and y?

Integrals with respect to x and y have many applications in mathematics, physics, engineering, and other fields. They can be used to find the area, volume, and centroid of complex shapes, calculate moments of inertia, and solve differential equations. They are also used in probability and statistics to find the probability of events and expected values.

## How can I improve my skills in solving integrals with respect to x and y?

To improve your skills in solving integrals with respect to x and y, it is important to have a strong understanding of basic calculus concepts such as differentiation and single-variable integration. Practice is also key, so consider solving a variety of problems and seeking help from textbooks, online resources, or a tutor if needed. Additionally, familiarizing yourself with various integration techniques and their applications can also be beneficial.

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