Volume obtained by rotating region around x-axis

[rir0302]

Find the volume of a solid obtained by rotating the region enclosed by the graphs y=19−2x, y=5, and x=0 about the x‑axis.

I'm not sure how to do this problem. Is it the washer method? What would the formula be?

 

[MarkFL]

I would begin by making a sketch of the region to be rotated:



We can use either the washer or shell methods here. Suppose we try the washer method. Can you identify the outer and inner radii of an arbitrary washer?

 

[rir0302]

Oh, would it be upper - lower because it rotates around the x-axis?
So would the outer radius be y=19-2x, and the inner radius would be y=5?

 

[MarkFL]

Oh, would it be upper - lower because it rotates around the x-axis?
So would the outer radius be y=19-2x, and the inner radius would be y=5?

Good, yes. So, the volume of an arbitrary washer would be:

[MATH]dV=\pi\left((19-2x)^2-(5)^2\right)\,dx[/MATH]
What would your limits of integration be?

 

[rir0302]

So the limits would be 5 to 7, I think... and then would we have to do anything else about the gap between the region and the x-axis?

 

[Deleted member 4993]

So the limits would be 5 to 7, I think... and then would we have to do anything else about the gap between the region and the x-axis?

Incorrect.

You are integrating wrt 'x'. So the limits should be 0 to 7.

In your expression for dV,

dV=π((19−2x)²−(5)²)dx

the -5² part is taking care of the gap. To see these clearly, you should draw a sketch of the washer.

 

[rir0302]

Oh I see, thank you!

 

[MarkFL]

We may also use the shell method:

[MATH]dV=2\pi y\left(\frac{19-y}{2}\right)\,dy=\pi(19y-y^2)\,dy[/MATH]
Hence:

[MATH]V=\pi\int_5^{19} 19y-y^2\,dy[/MATH]