Double Integral over a triangular region

[mman014]

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

∫$^{1}_{0}$∫$^{1}_{.5}$sin(∏*y^2) dxdy
Integrating with respect to x first gives
.5∫$^{1}_{0}$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.

 

[SammyS]

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

∫$^{1}_{0}$∫$^{1}_{.5}$sin(∏*y^2) dxdy
Integrating with respect to x first gives
.5∫$^{1}_{0}$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.

 

[mman014]

Thanks.

Anyways so I figured out that I had the limits wrong, and when I change it I am getting
∫$^{1}_{0}$∫$^{sin(∏*y^2)}_{0}$sin(∏*y^2)dxdy
So from there I get
∫$^{1}_{0}$sin(∏*y^2)[x]$^{sin(∏*y^2)}_{0}$
= ∫$^{1}_{0}$ sin$^{2}$(∏*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.

 

[SammyS]

Thanks.

Anyways so I figured out that I had the limits wrong, and when I change it I am getting
∫$^{1}_{0}$∫$^{sin(∏*y^2)}_{0}$sin(∏*y^2)dxdy
So from there I get
∫$^{1}_{0}$sin(∏*y^2)[x]$^{sin(∏*y^2)}_{0}$
= ∫$^{1}_{0}$ sin$^{2}$(∏*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."​

 

[mman014]

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.

 

[SammyS]

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 !

 

[mman014]

Anyways thanks for making me realize my mistake haha. I was staring at the problem for too long, fresh eyes always help.