Calculus I (AB): Volume with Known Cross-Sections

I define region S as the area bounded by , y = 0, and x = 4. If S is the base of a solid whose semicircular cross-sections are perpendicular to the x-axis, find the volume of that solid.

Difficulty:

Solution:

Let's start by taking a look at the region which will be the base of the solid:

Above, the blue rectangle represents one of the cross-sections. Remember, we're looking at the solid from a top-down view. The semicircle's diameter is the blue rectangle, and the semicircle is coming up out of your computer monitor, coming right at you (protect your eyes). The volume of a figure such as this (any known cross-section) is the integral of one of the cross-sections. Since the blue rectangle is vertical (perpendicular to the x-axis), everything must be in terms of x, which means our integral must contain all x's, including x boundaries on the integral. Clearly, the x-boundaries are 0 and 4, and the integral will be

This is because the area of a semicircle is the area of a circle divided by 2. Now, we need to find r, the radius of the semicircle, in terms of x. Since the blue rectangle above represents the diameter of a semicircular cross-section, we can divide its length by 2 to get the radius:

How do we know that the length of the diameter is ? Because its height is defined by the function, which is also . Now, plug r into the cross-section integral and solve:

After all that work, you're rewarded with pi. Tasty. Ain't math grand?

Determine whether or not the series converges.

Difficulty:

Solution:

Because both the numerator and denominator will be pretty big honking numbers with n gets large, it's best to use the Ratio Test to determine convergence. In the Ratio Test, you examine the n + 1 term in a series divided by the n term in the series. Specifically, you take the limit of that fraction as n approaches infinity. If the result is less than 1, the series converges; if it's greater than 1, then the series diverges. If the result is 0, the Ratio Test fails, and you have to try a different test:

Dividing two fractions is the equivalent of multiplying the top value by the reciprocal of the bottom, so that's what we'll do. Then, simplify the fraction completely.

Since this infinite limit is greater than 1, the series must diverge.