Calculus I (AB): Area Between Curves and Rotational Volume

I define region R as the area in the first quadrant bounded by y = cos x, y = sin x, and x = 0.

(a) Find the area of R.

(b) Calculate the volume generated by revolving R about the y-axis.

You may use a graphing calculator on part (b).

Difficulty:

Solution:

(a) Let's start by taking a gander at the graphs and the region they bound:

The blue graph is cosine, and the red is sine. The graph of x = 0 is the y-axis. Therefore, the light green area is Region R. The x value of the region's right edge is x = /4. To calculate the area of R, we integrate the top curve minus the bottom curve on the x interval defining its boundaries. (Remember that x = 0 represents the left edge of the region, since it is bounded by the y-axis.)

This is approximately equal to .414.

(b) The rotational shape will be solid, but the disk method is not a great idea, because you'd have to translate everything in terms of y. The shell method is a better idea. Since the region is right up against the rotational axis, d(x) = x in the shell method, and h(x), the height of the rotational rectangles, is top minus bottom (cos x - sin x) just like in part (a):

You can distribute the x, split this up into two integrals, and then integrate each parts if you like, but integration by parts is an advanced topic. I elect to use a computational device (in my case a graphing calculator) to find the answer, which is approximately .696.

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Find the sum of the infinite series .

Difficulty:

Solution:

Even if you didn't know this was a geometric series, you could easily see the pattern if you divide all the terms by the first term:

Therefore, this is a geometric series of form , with a = 3 and r = 2/3. Since , the series converges to the following sum: