Applications of Integration: Volumes Part I

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Hckyplayer8

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Sketch the region enclosed by the graphs of y=x2 and y=2x+3 and find the volume of the solid produced when that region is revolved about the x-axis.

I sketched the region which exists from the lower limit -1 and upper limit 3 and is bounded by the two functions.

I know based off the disk method, that volume equals the definite integral of (pi)(r2) in respect to x.

Being that the "lower function" is the base of my parabola which results in a radius that changes with x, what am I suppose to use as my radius for this problem?
 
Have you learned the "washer method"? Essentially, you subtract the volume inside the inner surface from the volume inside the outer surface. You'll use both radii in the integrand.
 
I have covered the washer method.

I'm trying to visualize this figure three dimensionally, but I'm failing. How is there a hole when the figure is revolved around the x axis considering the parabola literally touches the x axis at x=0?
 
Hckyplayer8 said:
I have covered the washer method.

I'm trying to visualize this figure three dimensionally, but I'm failing. How is there a hole when the figure is revolved around the x axis considering the parabola literally touches the x axis at x=0?
Click to expand...
The linked page tells you exactly what r is.
 
Hckyplayer8 said:
I have covered the washer method.

I'm trying to visualize this figure three dimensionally, but I'm failing. How is there a hole when the figure is revolved around the x axis considering the parabola literally touches the x axis at x=0?
Click to expand...
The hole does not have to be all the way through.

The idea is the EACH SLICE is a "washer" with a hole in the middle. The fact that one slice, right at the end, has no hole, doesn't matter; the inner radius just happens to be 0 there. For every slice, you will have an outer radius (the upper function) and an inner radius (the lower function).
 
Dr.Peterson said:
The hole does not have to be all the way through.

The idea is the EACH SLICE is a "washer" with a hole in the middle. The fact that one slice, right at the end, has no hole, doesn't matter; the inner radius just happens to be 0 there. For every slice, you will have an outer radius (the upper function) and an inner radius (the lower function).
Click to expand...

So if the solution of the problem is the difference of the two washers then the setup for this problem is
the definite integral over [-1,3] (pi)((2x+3)^2) - ((x^2)^2) dx
 
Hckyplayer8 said:
So if the solution of the problem is the difference of the two washers then the setup for this problem is
the definite integral over [-1,3] (pi)((2x+3)^2) - ((x^2)^2) dx
Click to expand...
Just to make sure you're saying what you mean, it is not "the difference of the two washers"; rather, the elements we are integrating are "washers", whose volumes are the difference between two cylinders (squaring the outside and inside radii).

But what you mean is correct.

Similarly, your work is almost correct, but the first few lines are missing parentheses, which could have led you astray if you did what you wrote; and the third integral is not at all what you mean, as at that point you have already integrated! There should be no integral sign at that point; you just need to show that you are evaluating from -1 to 3, which you carry out in the next step.

Rather, you should write the step you skipped, [MATH]\pi\int_{-1}^{3} (4x^2 + 12x + 9 - x^4)dx[/MATH]. I had to write this step out to see that you had not made a mistake in one of your numbers! So should you.
 
Dr.Peterson said:
Just to make sure you're saying what you mean, it is not "the difference of the two washers"; rather, the elements we are integrating are "washers", whose volumes are the difference between two cylinders (squaring the outside and inside radii).

But what you mean is correct.

Similarly, your work is almost correct, but the first few lines are missing parentheses, which could have led you astray if you did what you wrote; and the third integral is not at all what you mean, as at that point you have already integrated! There should be no integral sign at that point; you just need to show that you are evaluating from -1 to 3, which you carry out in the next step.

Rather, you should write the step you skipped, [MATH]\pi\int_{-1}^{3} (4x^2 + 12x + 9 - x^4)dx[/MATH]. I had to write this step out to see that you had not made a mistake in one of your numbers! So should you.
Click to expand...

Thank you for the feedback.

I understand dropping the integral symbol after integrating.

Which lines are missing parentheses? The first with the original equation and then the second with the functions plugged in? I also see I forgot to initially declare which variable I'm integrating by.
 
My final answer came out to be 1088pi over 15...which matches the generator on symbolab.

My book doesn't have any units of measure attached to it's answers. Would it be wrong to assume all of these volumes are cubic units?
 
Hckyplayer8 said:
My final answer came out to be 1088pi over 15...which matches the generator on symbolab.

My book doesn't have any units of measure attached to it's answers. Would it be wrong to assume all of these volumes are cubic units?
Click to expand...

If no units are specified, then your assumption is spot on. :)
 
Hckyplayer8 said:
Which lines are missing parentheses? The first with the original equation and then the second with the functions plugged in? I also see I forgot to initially declare which variable I'm integrating by.
Click to expand...
Right. As written, pi multiplies only the first term. I personally also prefer putting parentheses around the entire integrand, separating it from dx; but many people consider that to be implied.
 
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