Calculus III: Volume of an area limited by two curves

[MattDaCat]

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]

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?

 

[MattDaCat]

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]

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]

Yes, that all looks good to me.