The area of a region R is given by the iterated integral $ \int_{0}^{8} \int_{\frac{x}{8}}^{1} d y d x $ Switch the order of integration and show that both orders yield the same area. What is this area?
A) 65
B) 16
C) 21
D) 8
E) 4

Question

Quizplus · Accepted Answer

The given iterated integral is $\int_{0}^{8} \int_{\frac{x}{8}}^{1} d y d x$. To switch the order of integration, we first identify the limits for $x$ and $y$. From the given integral, $x$ ranges from 0 to 8, and $y$ ranges from $\frac{x}{8}$ to 1. To switch the order, we solve $\frac{x}{8}=y$ for $x$, giving $x=8y$. Thus, $x$ ranges from $0$ to $8y$, and since $y$ goes up to 1 in the original integral, we set the new limits for $y$ from its minimum value in the original integral, which is $\frac{x}{8}$ at $x=0$, to its maximum, which is 1. Therefore, the new integral becomes $\int_{0}^{1} \int_{0}^{8y} dx dy$. Evaluating this integral gives the area of the region:$$\int_{0}^{1} \int_{0}^{8y} dx dy = \int_{0}^{1} [8y]_{0}^{8y} dy = \int_{0}^{1} 8y dy = [4y^2]_{0}^{1} = 4.$$Thus, both orders of integration yield the same area, which is 4.

The Area of a Region R Is Given by the Iterated