MHB What are the y limits for finding the volume of a solid with given parameters?

[cbarker1]

Find the volume of the following solid:

The solid lies below the surface z=x^3y and above the triangle in the xy plane with vertices (1,0), (2,1) and (4,0).
The region is graphed
[desmos="-10,10,-10,10"](1,0);(4,0);(2,1);[/desmos]I need to find the y limits in the double integral.

Thanks

CBarker1

 

[I like Serena]

Find the volume of the following solid:

The solid lies below the surface z=x^3y and above the triangle in the xy plane with vertices (1,0), (2,1) and (4,0).
The region is graphed below:

I need to find the y limits in the double integral.

Thanks

CBarker1

Hi CBarker1!

I think that first equation should be $y=x-1$, otherwise it doesn't contain (2,1).

And there you have both your y limits.

The volume is:
$$V=\int_0^{x-1}\int_1^2 x^3y\,dx\,dy + \int_0^{-\frac 12 x+2}\int_2^4 x^3y\,dx\,dy$$

 

[cbarker1]

What if I integrate with respect to x for the second integral?

 

[I like Serena]

What if I integrate with respect to x for the second integral?

Then we have to invert those 2 functions, and we get
$$V=\int_{y+1}^{-2y+4}\int_0^1 x^3y\,dy\,dx$$

 

[cbarker1]

I need a value answer such as 3 units cubic. So something is wrong because that will give me a general formula. So the limits inverted? Because the constant limits should be at the outside integral.

 

[I like Serena]

I need a value answer such as 3 units cubic. So something is wrong because that will give me a general formula. So the limits inverted? Because the constant limits should be at the outside integral.

Yes. It should be:
$$V=\int_0^1 \int_{y+1}^{-2y+4} x^3y\,dx\,dy$$