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Double integrals help

  • #1
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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}##
 

Answers and Replies

  • #2
LCKurtz
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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 ##
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
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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
LCKurtz
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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
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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 theres 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
LCKurtz
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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 theres 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
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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}##
 

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