Double Integral Evaluation: Why Do We Integrate with Respect to y First?

[DryRun]

Homework Statement
Evaluate the double integral:
http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

The attempt at a solution

Consider x as constant, first integrate w.r.t.y.

I get
$$\frac{xy^2}{2}$$
After evaluating the integral with limits y=0 to y=y, i get the same exact answer.

Then, i integrate the above w.r.t.x

I get
$$\frac{x^2 y^2}{4}$$
Applying the limits x=0 to x=1

I get the final answer as:
$$\frac{y^2}{4}$$

However, the answer in my notes says: 1/8

Also, i don't understand why i must always first integrate w.r.t.y then w.r.t.x and not vice versa?

 

[Mark44]

Homework Statement
Evaluate the double integral:
http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

The attempt at a solution

Consider x as constant, first integrate w.r.t.y.

No. You need to integrate with respect to x first. That is the significance of dx coming before dy.

I get
$$\frac{xy^2}{2}$$
After evaluating the integral with limits y=0 to y=y, i get the same exact answer.

Then, i integrate the above w.r.t.x

I get
$$\frac{x^2 y^2}{4}$$
Applying the limits x=0 to x=1

I get the final answer as:
$$\frac{y^2}{4}$$

However, the answer in my notes says: 1/8

Also, i don't understand why i must always first integrate w.r.t.y then w.r.t.x and not vice versa?

It depends on the order of dx and dy in the iterated integral.
In this integral you integrate with respect to x first, and then y.
$$\int \int f(x,y) dx~dy$$

Here you integrate with respect to y first, and then x.
$$\int \int f(x,y) dy~dx$$

 

[DryRun]

Hi Mark44

Using your method, i get the final answer:
$$\frac{x^2}{4}$$
which is wrong.

OK, for argument's sake (and to clear up my confusion), let's just consider another example:
http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

From what you tell me, the answer should be: 2-2x
Again, this is wrong. So, something about your method doesn't add up.

 

[ehild]

Hi Mark44

Using your method, i get the final answer:
$$\frac{x^2}{4}$$
which is wrong.

How did you get that final answer? It has to be a number. Have you substituted the upper and lower bounds after integration with respect to x?

ehild

 

[Curious3141]

Hi Mark44

Using your method, i get the final answer:
$$\frac{x^2}{4}$$
which is wrong.

The answer should be numerical. First integrate wrt x, treating y as a constant. Then evaluate the definite integral between the bounds 0 and y, which means substituting y for x in that expression to get a form that only has y in it. Finally integrate that expression wrt y with the bounds 0,1.

OK, for argument's sake (and to clear up my confusion), let's just consider another example:
http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

From what you tell me, the answer should be: 2-2x
Again, this is wrong. So, something about your method doesn't add up.

As it is written, the answer is 2 - 2x. This is because the inner bound with the sqrt(x) leaves the variable x in the integral after integrating wrt x and evaluating for the bounds, and it persists after the final integration wrt y.

 

[DryRun]

OK, i got the correct numerical answer for the first problem:
http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

However, for the second problem:
http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg
The answer in my notes is numerical: 6/7
The solution (according to my notes): The variable x is considered as constant and then integrate w.r.t.y (which is the other way round to what I've been led to believe in this thread!). I'm confused.

 

[gb7nash]

The problem looks like it was copied or written wrong, then. It doesn't make sense to integrate with respect to x, and then plug in a function of x as a bound. My guess (and you'll have to check with your instructor) is that the second problem is supposed to be:

$$\int_0^1 \int_\sqrt{x}^1 12xy^2 dy dx$$

 

[jackmell]

I'm confused.

I think you have a problem with the geometry so I believe you should focus on it for now rather than the algebra: Suppose I said I want the volume underneath the function $f(x,y)=12 x^2$ that lies between the curves $y_1(x)=\sqrt{x}$ and $y_2(x)=1$ from the points x=0 to x=1. Now forget about the integrals for now. Just draw that, nicely. Label everything. Now go back to your text and study how the double integral is defined as either:

$$\int_{x_1}^{x_2}\int_{y_1(x)}^{y_2(x)} f(x,y)dydx$$

or if instead I have the functions in terms of y as $x_1(y)$ and $x_2(y)$ from the points $y_1$ to $y_2$:

$$\int_{y_1}^{y_2}\int_{x_1(y)}^{x_2(y)} f(x,y)dxdy$$

and then adapt your problem to one of those.

 

[Mark44]

The problem looks like it was copied or written wrong, then. It doesn't make sense to integrate with respect to x, and then plug in a function of x as a bound. My guess (and you'll have to check with your instructor) is that the second problem is supposed to be:

$$\int_0^1 \int_\sqrt{x}^1 12xy^2 dy dx$$

I agree.

 

[SammyS]

The problem looks like it was copied or written wrong, then. It doesn't make sense to integrate with respect to x, and then plug in a function of x as a bound. My guess (and you'll have to check with your instructor) is that the second problem is supposed to be:

$$\int_0^1 \int_\sqrt{x}^1 12xy^2 dy dx$$

Of course, evaluating this does not give the result 6/7.

 

[DryRun]

OK, thank you all for your help. I think it is a typo error then. I will email my instructor.

Best wishes for the holidays!

 

[Harrisonized]

The problem looks like it was copied or written wrong, then. It doesn't make sense to integrate with respect to x, and then plug in a function of x as a bound. My guess (and you'll have to check with your instructor) is that the second problem is supposed to be:

http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

Actually, I think this was an intentional error. If you draw the region, you'll find that you can change the integral to

[STRIKE]∫₀¹∫₀^{y^2} 12xy² dx dy

Evaluation of this integral DOES result in 6/7.[/STRIKE]

 

[DryRun]

I'm struggling to understand the basic concepts here. Without a solid foundation, i can't proceed with further sections in double integrals and i have to study triple integrals after that. Can someone please give me a definitive answer?

 

[Harrisonized]

Sorry. I made a mistake earlier. It should be

∫₀¹∫₀^√x 12xy² dy dx

which results in 8/7.

Whenever you have a double integral, you should just treat it as an integral within an integral.

∫₀¹∫_√x¹ 12xy² dx dy

= ∫₀¹ (∫_√x¹ 12xy² dx) dy

As with a regular integral:

∫_a^b f'(x,y) dx

a represents the lower bound and b represents the upper bound with respect to x. In the case of general regions, instead of having a line as the lower bound, the lower bound is, instead, a curve.

Whenever there is a multiple integrals over a general region, you want all the variables to disappear by the end of the integration. In other words, when f(x) is one of the bounds on an integral, there had better be a dx on one of the outer layer of the nested integral.

In your case, the bounds with respect to x is √x to 1. The only outer layer is an integral with respect to y. This is a problem, since integrating outside with respect to y does not get rid of the function of x in the x-bound. The only choice here is to change how the region of integration is defined.

If you draw the region on a piece of paper, you'll find that the region of integration is the inner section of the upper region of a horizontal parabola, with vertex at (0,0), cut off at the line x=1. One easy way rewrite the bounds is to have the y run from 0 to √x. Then the x bounds run from 0 to 1. This means that you'll be integrating with respect to y on the inner layer of the integral, and you'll be integrating with respect to x on the outer layer.

∫₀¹∫₀^√x 12xy² dy dx

= ∫₀¹(∫₀^√x 12xy² dy) dx

= ∫₀¹ (4xy³ |_y=0^y=√x) dx

= ∫₀¹ (4xx^3/2) dx

= ∫₀¹ 4x^5/2 dx

= 4 (2/7) x^7/2 |₀¹

= 8/7

Keep in mind that the integral you wrote down:

http://s1.ipicture.ru/uploads/20111224/VxWfVW7H.jpg

doesn't actually make sense by itself. My interpretation is just one way to crunch numerical results out of the nonsensical integral.

 

[DryRun]

OK, I'm beginning to see a clearer pathway of reasoning. However, the first problem appears to be a trick question, since it has 2 possible answers??

 

[Mark44]

OK, I'm beginning to see a clearer pathway of reasoning. However, the first problem appears to be a trick question, since it has 2 possible answers??

No, there is just one answer -- 1/8, the same as in your notes.

After the first (inner) integration, you should get y³/2. After integrating that with respect to y, you should get y⁴/8, which you evaluate at 1 and at 0, resulting in the final answer of 1/8.

Homework Statement
Evaluate the double integral:
http://s1.ipicture.ru/uploads/20111224/K3MGcdS7.jpg

The attempt at a solution

Consider x as constant, first integrate w.r.t.y.

I get
$$\frac{xy^2}{2}$$
After evaluating the integral with limits y=0 to y=y, i get the same exact answer.

Then, i integrate the above w.r.t.x

I get
$$\frac{x^2 y^2}{4}$$
Applying the limits x=0 to x=1

I get the final answer as:
$$\frac{y^2}{4}$$

However, the answer in my notes says: 1/8

Also, i don't understand why i must always first integrate w.r.t.y then w.r.t.x and not vice versa?

 

[DryRun]

OK, i got it. Thanks to all.