Shell method with region in first quadrant only

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rir0302

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I have to use the shell method to find the volume between y=-x+9 and y=2x-3 but only the region in the first quadrant.
So the graph would look like this:
Screen Shot 2019-09-23 at 5.14.18 PM.png

I set up the integral as 2π∫[x((-x+9)-(2x-3))]dx from [0, 4] but I think that would also include that triangle region that's not in the first quadrant?
How would you write the integral so that it doesn't?
 
One way would be to do the integral you wrote, and then subtract the area of the triangle.

Another would be to split the integral into two parts, one for the region where 2x-3 is below the axis (so your integrand is just -x+9), and another where both lines are above the axis (using the integrand you showed).

Or, you could integrate with respect to y, but that would also need two parts.
 
Your attached image isn't working for me...here's a diagram of the region:

fmh_0082.png

What is the axis of rotation?
 
Oh sorry, does this work?:

Graph.png

So, could I use the volume of the entire region and subtract the volume of the non-included region from it?
Something like 2π∫[x((-x+9)-(2x-3))]dx [0, 4] - 2π∫[x(0-(2x-3)]dx [0, 1.5] (I'm not sure if the second integral is right)
 
Please refer to this diagram:

fmh_0083.png

For the area shaded in red:

[MATH]V_1=2\pi\int_0^{\frac{3}{2}} x(9-x)\,dx=2\pi\int_0^{\frac{3}{2}} 9x-x^2\,dx[/MATH]
And for the area shaded in green:

[MATH]V_2=2\pi\int_{\frac{3}{2}}^4 x((9-x)-(2x-3))\,dx=6\pi\int_{\frac{3}{2}}^4 4x-x^2\,dx[/MATH]
And then:

[MATH]V=V_1+V_2[/MATH]
 
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