M MattDaCat New member Joined Feb 16, 2020 Messages 2 Feb 16, 2020 #1 The problem is as follows: [MATH] \iint_D (12-x)y \,dx\,dy [/MATH] Where D is the area limited by: 12-4y-x=0 and x=y2 I'm new to the course so any help is appreciated!
The problem is as follows: [MATH] \iint_D (12-x)y \,dx\,dy [/MATH] Where D is the area limited by: 12-4y-x=0 and x=y2 I'm new to the course so any help is appreciated!
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Feb 16, 2020 #2 Hello, and welcome to FMH! I would begin with a diagram of the region over which we'll be integrating: Would you identify this as a type I or type II region?
Hello, and welcome to FMH! I would begin with a diagram of the region over which we'll be integrating: Would you identify this as a type I or type II region?
M MattDaCat New member Joined Feb 16, 2020 Messages 2 Feb 16, 2020 #3 I would say this is a type II region, bounded by the lines y=-6 and y=4. Putting x as a function of y: x = y2 x = 12 - 4y Would this integral give the correct answer? [MATH] \int_{-6}^{4} \int_{y^2}^{12-4y} (12-x)y \,dx\,dy [/MATH]
I would say this is a type II region, bounded by the lines y=-6 and y=4. Putting x as a function of y: x = y2 x = 12 - 4y Would this integral give the correct answer? [MATH] \int_{-6}^{4} \int_{y^2}^{12-4y} (12-x)y \,dx\,dy [/MATH]
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Feb 16, 2020 #4 MattDaCat said: I would say this is a type II region, bounded by the lines y=-6 and y=4. Putting x as a function of y: x = y2 x = 12 - 4y Would this integral give the correct answer? [MATH] \int_{-6}^{4} \int_{y^2}^{12-4y} (12-x)y \,dx\,dy [/MATH] Click to expand... Yes, that all looks good to me.
MattDaCat said: I would say this is a type II region, bounded by the lines y=-6 and y=4. Putting x as a function of y: x = y2 x = 12 - 4y Would this integral give the correct answer? [MATH] \int_{-6}^{4} \int_{y^2}^{12-4y} (12-x)y \,dx\,dy [/MATH] Click to expand... Yes, that all looks good to me.
