Am I on the right track for this problem?

[goku900]

\(\displaystyle \mbox{14. Eval}\mbox{uate the inte}\mbox{gral:}\)

\(\displaystyle \displaystyle{\iint\limits_R}\, (y^2\, +\, xy\, -\, 2x^2)\, dA\)

\(\displaystyle \mbox{where }\, R\, \mbox{ is the region bounded by the lines }\, y\, =\, x,\)

\(\displaystyle y\, =\, x\, -\, 3,\, y\, =\, -2x\, +\, 3,\, y\, =\, -2x\, -\, 3\)

In this problem I have drawn out the region specified and noticed two sets of parallel lines indicating to me that a change of variable(u and v) are able to be used to solve this integral.

I decided that u=y-x and v = -2x-y then solving for x and y I obtain x= (u-v)/3 and y = (4u-v)/3

From here I can compute the jacobian but I want to be sure the work above is correct before I move on. Can anyone please confirm?

 

[Romsek]

\(\displaystyle \mbox{14. Eval}\mbox{uate the inte}\mbox{gral:}\)

\(\displaystyle \displaystyle{\iint\limits_R}\, (y^2\, +\, xy\, -\, 2x^2)\, dA\)

\(\displaystyle \mbox{where }\, R\, \mbox{ is the region bounded by the lines }\, y\, =\, x,\)

\(\displaystyle y\, =\, x\, -\, 3,\, y\, =\, -2x\, +\, 3,\, y\, =\, -2x\, -\, 3\)

In this problem I have drawn out the region specified and noticed two sets of parallel lines indicating to me that a change of variable(u and v) are able to be used to solve this integral.

I decided that u=y-x and v = -2x-y then solving for x and y I obtain x= (u-v)/3 and y = (4u-v)/3

From here I can compute the jacobian but I want to be sure the work above is correct before I move on. Can anyone please confirm?

View attachment 3460

How to solve this really depends on what you are studying at the moment.

If you are studying the brute force way to evaluate double integrals you need to observe that there are 3 areas that have to be treated separately,

x=(-1,0), x=(0,1), x=(1,2).

The limits of integration for y are different in these 3 areas. Can you see why?

For each of these 3 areas you have the function of x that are the upper and lower limits for y. So for example for the first area you get

\(\displaystyle \int _{-1}^0\int _{-2 x-3}^x\left(-2 x^2+x y+y^2\right)dydx\)

You proceed similarly for the other 2 areas. The final value of the integral is the sum of the 3 integrals.

So I don't really see any need for the substitution you made.

If you're studying Green's theorem you'd approach it entirely differently, so I need to know what it is your are currently studying.

 

[goku900]

Ok im going to sovle for the second two

 

[goku900]

View attachment 3460

How to solve this really depends on what you are studying at the moment.

If you are studying the brute force way to evaluate double integrals you need to observe that there are 3 areas that have to be treated separately,

x=(-1,0), x=(0,1), x=(1,2).

The limits of integration for y are different in these 3 areas. Can you see why?

For each of these 3 areas you have the function of x that are the upper and lower limits for y. So for example for the first area you get

\(\displaystyle \int _{-1}^0\int _{-2 x-3}^x\left(-2 x^2+x y+y^2\right)dydx\)

You proceed similarly for the other 2 areas. The final value of the integral is the sum of the 3 integrals.

So I don't really see any need for the substitution you made.

If you're studying Green's theorem you'd approach it entirely differently, so I need to know what it is your are currently studying.[/QUOTE]

So the 2nd and 3rd would look like this?

\(\displaystyle \int _{0}^1\int _{x-3}^x\left(-2 x^2+x y+y^2\right)dydx\)

\(\displaystyle \int _{1}^2\int _{x-3}^{2x+3}\left(-2 x^2+x y+y^2\right)dydx\)

 

[Romsek]

So the 2nd and 3rd would look like this?

\(\displaystyle \int _{0}^1\int _{x-3}^x\left(-2 x^2+x y+y^2\right)dydx\)

\(\displaystyle \int _{1}^2\int _{x-3}^{2x+3}\left(-2 x^2+x y+y^2\right)dydx\)

yes, exactly right.

my mistake the 3rd integral should have upper limit of -2x+3 as Daon noted.

 

[goku900]

yes, exactly right.

The area comes out negative when computed this way

 

[daon2]

The area comes out negative when computed this way

your third integral should have an upper bound of -2x+3, also you are not finding an area (when the integrand is just "1"), you are finding the integral of your function F(x,y)

 

[goku900]

your third integral should have an upper bound of -2x+3, also you are not finding an area (when the integrand is just "1"), you are finding the integral of your function F(x,y)

ok so I get 27/8 + 18/8 -45/8 = 0 as an answer using those three integrals in order