Double Integral over a triangular region

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Homework Help Overview

The discussion revolves around calculating the volume under the surface defined by the function f(x,y) = sin(∏*y^2) over a triangular region R in the x-y plane, specifically defined by the vertices (0,0), (0,1), and (.5,1). Participants are exploring the use of double integrals to solve this problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of the double integral and the limits of integration, with initial attempts leading to confusion over the correct boundaries. There are considerations of reversing the order of integration and the challenges posed by the lack of a simple anti-derivative for the function involved. Some participants also explore the implications of changing the limits and the potential use of trigonometric substitution.

Discussion Status

The discussion has progressed with participants identifying errors in their initial setups and adjusting their approaches. Some have expressed uncertainty about their limits and integration methods, while others have suggested writing equations for the boundaries of the triangular region to clarify the problem. There is a recognition of mistakes, and a participant has indicated a clearer understanding after reevaluating their approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the methods they can use or the depth of exploration allowed. The discussion reflects a collaborative effort to clarify the problem setup and integration techniques without providing direct solutions.

mman014
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Homework Statement


Let f(x,y) = sin(∏*y^2). Let R be the triangular region on the x-y plane with the vertices at (0,0) (0,1) (.5,1). Consider the solid that is under z = f(x,y) and over the region R. Calculate the volume over that region using double integrals.

Homework Equations





The Attempt at a Solution


∫[itex]^{1}_{0}[/itex]∫[itex]^{1}_{.5}[/itex]sin(∏*y^2) dxdy
Integrating with respect to x first gives
.5∫[itex]^{1}_{0}[/itex]sin(∏*y^2)dy.
After this I'm stuck because I know there is no simple anti-derivative for that function.
If you reverse the order of integration then, sin(∏*y^2) would have to be integrated at first as well, so not sure what to do.
 
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mman014 said:

Homework Statement


Let f(x,y) = sin(∏*y^2). Let R be the triangular region on the x-y plane with the vertices at (0,0) (0,1) (.5,1). Consider the solid that is under z = f(x,y) and over the region R. Calculate the volume over that region using double integrals.

Homework Equations



The Attempt at a Solution


∫[itex]^{1}_{0}[/itex]∫[itex]^{1}_{.5}[/itex]sin(∏*y^2) dxdy
Integrating with respect to x first gives
.5∫[itex]^{1}_{0}[/itex]sin(∏*y^2)dy.
After this I'm stuck because I know there is no simple anti-derivative for that function.
If you reverse the order of integration then, sin(∏*y^2) would have to be integrated at first as well, so not sure what to do.
Hello mman014. Welcome to PF !

You are integrating over a rectangular region.

Write an equation for each line that is on the boundary of region R.
 
Thanks.

Anyways so I figured out that I had the limits wrong, and when I change it I am getting
∫[itex]^{1}_{0}[/itex]∫[itex]^{sin(∏*y^2)}_{0}[/itex]sin(∏*y^2)dxdy
So from there I get
∫[itex]^{1}_{0}[/itex]sin(∏*y^2)[x][itex]^{sin(∏*y^2)}_{0}[/itex]
= ∫[itex]^{1}_{0}[/itex] sin[itex]^{2}[/itex](∏*y^2) dy
At this point I am stuck again, unless I am doing something wrong or I found the wrong limits again. From here the only way I can think of continuing is using trig substitution and then I would have
the integral of 1/2 - (cos2(∏*y^2))/2. Which doesn't really help because I would still have to integrate the cos(2∏2y^2) which is the same problem as before.
 
Last edited:
mman014 said:
Thanks.

Anyways so I figured out that I had the limits wrong, and when I change it I am getting
∫[itex]^{1}_{0}[/itex]∫[itex]^{sin(∏*y^2)}_{0}[/itex]sin(∏*y^2)dxdy
So from there I get
∫[itex]^{1}_{0}[/itex]sin(∏*y^2)[x][itex]^{sin(∏*y^2)}_{0}[/itex]
= ∫[itex]^{1}_{0}[/itex] sin[itex]^{2}[/itex](∏*y^2) dy
At this point I am stuck again, unless I am doing something wrong or I found the wrong limits again. From here the only way I can think of continuing is using trig substitution and then I would have
the integral of 1/2 - (cos2(∏*y^2))/2. Which doesn't really help because I would still have to integrate the cos(2∏2y^2) which is the same problem as before.

You continue to have them wrong.

As I suggested:
"Write an equation for each line that is on the boundary of region R."​
 
Wow so I just realized the mistake I made, wasn't thinking properly., equation of the line is just y = 2x so x = .5y, and I got it from there just using u sub and the rest is easy.
 
mman014 said:
Wow so I just realized the mistake I made, wasn't thinking properly., equation of the line is just y = 2x so x = .5y, and I got it from there just using u sub and the rest is easy.
Excellent !
 
Anyways thanks for making me realize my mistake haha. I was staring at the problem for too long, fresh eyes always help.
 

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