Post

Created by @nathanedwards
 at November 11th 2023, 8:13:40 pm.

Sure! Here's an AP Calculus AB exam question on the topic of volume of solids of revolution:

Question: A region is bounded by the curve y = x^2, the x-axis, and the lines x = 1 and x = 2. This region is rotated about the y-axis. a) Set up, but do not evaluate, the definite integral that represents the volume of the resulting solid. b) Evaluate the definite integral to find the volume of the solid.

Answer:

a) To set up the definite integral for the volume of the solid, we'll use the method of disks. For each vertical strip of the region, the radius of the resulting disk is given by the value of x, and the height (or thickness) is given by the difference between the x-coordinates of the bounds of integration.

The definite integral for the volume, V, of the solid is given by:

V=π12(x2)2dx V = \pi \int_{1}^{2} (x^2)^2 \, dx

b) To evaluate the definite integral, we need to find the antiderivative of x4 x^4 with respect to x:

x4dx=x55+C \int x^4 \, dx = \frac{x^5}{5} + C

Using the Fundamental Theorem of Calculus, we then evaluate the definite integral:

V=π[x55]12 V = \pi \left[ \frac{x^5}{5} \right]_{1}^{2}
V=π(255155) V = \pi \left( \frac{2^5}{5} - \frac{1^5}{5} \right)
V=π(32515) V = \pi \left( \frac{32}{5} - \frac{1}{5} \right)
V=31π5 V = \frac{31\pi}{5}

Therefore, the volume of the resulting solid is 31π5 \frac{31\pi}{5} cubic units.