Finding Volume of Solid Rotated X-Axis

[Cheapo2004]

Ok, I'm supposed to found the volume of the solid that is created after rotating the line f(x) = 2x-1 around the x axis. The limits are y=0 x=3 and x=0. I've been trying for about and hour, and keep getting the answer: 46.0766. I've done the integration tons of times, splitting the problem into two parts for each separate cone, and other stuff. I just can't seem to get the right answer, please help.

 

[marcin w]

Tell us how you set up the integral.

 

[tacman]

Hi there,

Finding the volume of a solid rotated around the x-axis can be challenging, but with the right approach, it is definitely possible. Let's go through the steps together.

First, we need to visualize the solid that is being rotated. In this case, we have a line f(x) = 2x-1, which looks like a straight line with a y-intercept of -1 and a slope of 2. When we rotate this line around the x-axis, we will get a shape that looks like a cone.

Next, we need to determine the limits of integration. Since we are rotating around the x-axis, the limits will be from x=0 to x=3. This means that we are only interested in the portion of the solid that lies between these two values of x.

Now, we need to set up the integral that will give us the volume of the solid. We can use the formula V = π∫(f(x))^2 dx, where f(x) is the function that defines the shape of the solid and dx is the infinitesimal thickness of the slices that we will be adding up.

Since we are rotating a line around the x-axis, the function f(x) will be the distance from the x-axis to the line at any given value of x. In this case, the distance is simply the value of f(x), which is 2x-1. So, our integral becomes V = π∫(2x-1)^2 dx.

Now, we need to integrate this function from x=0 to x=3. This will give us the volume of the solid between these two values of x. When we do the integration, we get V = π(64/3 - 1/3) = 63π/3 = 66.15.

So, the volume of the solid rotated around the x-axis is approximately 66.15 cubic units. This is different from the answer you got, which was 46.0766. Perhaps there was a mistake in your integration or in setting up the integral. I would recommend double-checking your work to see if you can find where the error occurred.

I hope this helps. Keep practicing and don't get discouraged. Integration can be tricky, but with enough practice, you will get the hang of it. Good luck!