Question

(Calculus) A fancy notebook has edges that follow the graphs of y = 1/2 sin(x) - 2 and y = 2 - 1/2 sin(x). The notepad is 14 inches long and 1 inch thick. Explain how you could use cross sections or slices perpendicular to the x-axis to find the notebook volume.

I get it that I take the integral of the cross-sectional area. Problem is, I'm geometrically challenged and don't know HOW to get that cross-section exactly. I have dimensions AND two functions, and this is where I'm getting confused. Can you guys walk me through the process and explain your steps?

Answer 1

Is there a picture associated with this? Having a hard time what they're calling "length" and "width" and the two answers would differ pretty wildly...

If the "length" of the notebook is 14" and the width is defined as the boundary between two functions, then you have a maximum width at pi/2. For y1 = 1/2 sin (x) - 2, this value is equal to 1/2 * 1 - 2 = -3/2, and at y2 = 2 - 1/2 sin (x) we have 2 - 1/2 = 3/2. So the maximum width of this notebook is going to be the distance between 3/2 and -3/2, or 6/2 = 3--so you've just bought yourself a fancy notebook that's 14" long and a maximum of 3" wide, money well spent.

Suspending disbelief, you can then integrate these two functions along x = [0, 14] and find the difference--note that sin(x) is bounded between (-1, 1) so 2 - 1/2 sin (x) will be bigger than 1/2 sin (x) - 2 for all x, so you would find the integral from 0 to 14 of 2 - 1/2 sin (x) - (1/2sin(x) - 2). This can be simplified to 2 - sin(x) + 2 = 4 - sin(x) from 0 to 14.

Next, you have the thickness is 1", so you multiply this cross section by its depth, i.e. 1.

I hope that clears things up for you a little bit. In a world of poorly worded cross-sectional area word problems, this one is right up there in the big leagues, but the important key to keep in mind with any cross-sectional area problem is that your shape is almost always just going to be a rectangle or a cylinder with a funny-shaped edge. In this case, we have a rectangular box with two funny-shaped edges, which can be dealt with simply enough.

Answer 2

No picture, I'm afraid... That was my first thought as well. You explained it very well. :) Thank you so much! I guessed I'd get the width by calculating the area between functions, but when it came to the thickness or determining integration boundaries, I lost it.

Answer 3

The way I figured it, if they meant that the total maximum length was 14" then the constants in your boundary functions (the +2 and -2) would be totally irrelevant, so I went with a sort of noodle-shaped notebook.

The "thickness" is always going to be straight back into your page. Whenever you do a cross-sectional area problem, your object is going to have to go straight back into the page--otherwise you would need ugly, horrible things from multivariable calculus to evaluate the volume. So at the end of what you're doing, you're always just going to be multiplying what's on your paper by some "thickness" constant.

Answer 4

Ah, I see what you mean. Makes sense considering that V = int[A(x)]dx when we use the height as a boundary variable... Am I right to be thinking of this in the sense that, since the thickness is a non-boundary variable, we just factor it out as a constant a la V = int[area between functions * thickness] = thickness * int[area between functions] from x = 0 to defined length?

Answer 5

whoops. Length, not height. -_-