Mastering Double Integrals: Solving Problems with Ease

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In summary, the double integral over the region y=x+3; y=x^2+1 is ##\int_{1}^{2}\int_{-\sqrt{y-1}}^{\sqrt{y-1}}dxdy + \int_{2}^{5}\int_{y-3}^{\sqrt{y-1}}dxdy = \frac{9}{2}##, with one integral in the dydx order and the other in the dxdy order.
  • #1
jonroberts74
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Homework Statement



Set up the double integral over the region ##y=x+3; y=x^2+1##

Homework Equations





The Attempt at a Solution



finding the intersections you get the double integral

##\int_{1}^{5}\int_{-1}^{2}dxdy =12 ##


but why is that not the same as

##\int_{1}^{5}\int_{-1}^{\sqrt{y-1}} dxdy =\frac{28}{3}##
 
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  • #2
jonroberts74 said:

Homework Statement



Set up the double integral over the region ##y=x+3; y=x^2+1##

Homework Equations





The Attempt at a Solution



finding the intersections you get the double integral

##\int_{1}^{5}\int_{-1}^{2}dxdy =12 ##

That would be the integral over a rectangle.

but why is that not the same as

##\int_{1}^{5}\int_{-1}^{\sqrt{y-1}} dxdy =\frac{28}{3}##

That's wrong too. Have you drawn the graphs?
 
  • #3
yeah I drew it, its the line y=x+3 slicing through the parabola x^2 moved up one unit on the y-axis

##\int_{1}^{5}\int_{-\sqrt{y-1}}^{\sqrt{y-1}} dxdy ## is the other one I had, but no sure if that is correct either
 
  • #4
jonroberts74 said:
yeah I drew it, its the line y=x+3 slicing through the parabola x^2 moved up one unit on the y-axis

##\int_{1}^{5}\int_{-\sqrt{y-1}}^{\sqrt{y-1}} dxdy ## is the other one I had, but no sure if that is correct either

It isn't. If you do x first it must go from ##x_{left}## to ##x_{right}##. x on the left is a two piece formula.

Is there some reason you don't do this problem the more natural and easier dydx order?
 
  • #5
LCKurtz said:
It isn't. If you do x first it must go from ##x_{left}## to ##x_{right}##. x on the left is a two piece formula.

Is there some reason you don't do this problem the more natural and easier dydx order?

probably out of habit of typically slicing with a y value then seeing how x changes

##\displaystyle\int_{-1}^{2}\int_{x^2+1}^{x+3} dydx##

okay I see it now, if I do x first there's an issue where the left intersection point occurs, below that point the parabola is the left and right x and above that the left is the line and the right is the parabola.

thanks!
 
  • #6
jonroberts74 said:
probably out of habit of typically slicing with a y value then seeing how x changes

##\displaystyle\int_{-1}^{2}\int_{x^2+1}^{x+3} dydx##

okay I see it now, if I do x first there's an issue where the left intersection point occurs, below that point the parabola is the left and right x and above that the left is the line and the right is the parabola.

thanks!

Yes! That's why you need two integrals to do it it dxdy order. That's the sort of thing you should always look at when deciding how to set up a double integral. Now can you get the same answer both ways?
 
  • #7
LCKurtz said:
Yes! That's why you need two integrals to do it it dxdy order. That's the sort of thing you should always look at when deciding how to set up a double integral. Now can you get the same answer both ways?
##\displaystyle \int_{1}^{2} \int_{-\sqrt{y-1}}^{\sqrt{y-1}}dxdy + \int_{2}^{5} \int_{y-3}^{\sqrt{y-1}}dxdy = \frac{4}{3} + \frac{19}{6} = \frac{9}{2}##
 

1. What is a double integral and why is it useful?

A double integral is a mathematical concept that involves calculating the area under a two-dimensional surface. It is useful because it allows us to find the volume or area of irregular shapes that cannot be easily calculated using traditional methods.

2. How do I set up and solve a double integral?

To set up a double integral, you need to define the boundaries of the region you want to find the area or volume of, and then choose the appropriate function to integrate over that region. To solve the double integral, you can use various techniques such as iterated integrals, change of variables, or polar coordinates.

3. What are some real-world applications of double integrals?

Double integrals are used in many fields such as physics, engineering, economics, and computer graphics. Some examples include calculating the center of mass of an object, finding the work done by a force, determining the area of a region on a map, and creating 3D models of objects.

4. What is the difference between a double integral and a single integral?

A single integral involves calculating the area under a one-dimensional curve, while a double integral involves calculating the area under a two-dimensional surface. This means that a single integral has one variable, while a double integral has two variables.

5. Are there any common mistakes to avoid when working with double integrals?

One common mistake is forgetting to properly define the boundaries of the region to be integrated. Another is not taking into account the orientation of the region, which can lead to incorrect results. It is also important to choose the correct order of integration, as switching the order can change the final result.

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