Setting up integrals for the area.

Question

Let R be the area in the first quadrant to the right of x = 1, above y = 2x^2, and below y = 6 + x.

A) Set up an integral that would find the volume of the region R rotated about the x-axis.

B) Set up an integral that would find the volume of the region R rotated about the y-axis.

You do not need to evaluate the integrals, just set them up, including bounds of integration.

Answer 1

Draw the region. The applicable region lies entirely in the first quadrant. In the first quadrant, y = 2x² and y = x+6 intersect at the point (2,8).

A. The upper boundary is the graph of y = x+6 and the lower boundary is the graph of y = 2x². Using the washer method, Volume = π∫_{(1 to 2)} [(x+6)² - (2x²)²]dx.

B. If y = 2x², then x = √(y/2). If y = x+6, then x = y - 6.

Between y = 0 and y = 6, the right boundary is the graph of x = √(y/2) and the left boundary is the graph

of x = 1 Between y = 6 and y = 8, the right boundary is the graph of x = √(y/2)and the right boundary is the graph of x = y-6 Using the washer method,

Volume = π∫_{(0 to 6)} [(√(y/2))² - 1²]dy + π∫_{(6 to 8)} [(√(y/2))² - (y-6)²]dy.