Set up $\iint _ { R } f ( x , y ) d A$ as an iterated integral (or more, if necessary) where you integrate first with respect to $y$ , where $R = \{ ( x , y ) : 0 \leq x \leq 2 \text { and } 0 \leq \mathrm { y } \leq 2 x \}$
A) $\int _ { 0 } ^ { 2 } \int _ { 2 x } ^ { 2 } f ( x , y ) d y d x$
B) $\int_{-2}^{2 x^{2}} \int_{-2}^{{+2}}f(x, y) d y d x$

C) $\int _ { 0 } ^ { 2 } \int _ { - \sqrt { 4 - x ^ { 2 } } } ^ { \sqrt { 4 - x ^ { 2 } } } f ( x , y ) d y d x$
D) $\int _ { - 2 } ^ { 2 } \int _ { - \sqrt { 4 - x ^ { 2 } } } ^ { \sqrt { 4 - x ^ { 2 } } } f ( x , y ) d y d x$

Question

Set up $\iint _ { R } f ( x , y ) d A$ as an iterated integral (or more, if necessary) where you integrate first with respect to $y$ , where $R = \{ ( x , y ) : 0 \leq x \leq 2 \text { and } 0 \leq \mathrm { y } \leq 2 x \}$
A) $\int _ { 0 } ^ { 2 } \int _ { 2 x } ^ { 2 } f ( x , y ) d y d x$
B) $\int_{-2}^{2 x^{2}} \int_{-2}^{{+2}}f(x, y) d y d x$

C) $\int _ { 0 } ^ { 2 } \int _ { - \sqrt { 4 - x ^ { 2 } } } ^ { \sqrt { 4 - x ^ { 2 } } } f ( x , y ) d y d x$
D) $\int _ { - 2 } ^ { 2 } \int _ { - \sqrt { 4 - x ^ { 2 } } } ^ { \sqrt { 4 - x ^ { 2 } } } f ( x , y ) d y d x$

Quizplus · Accepted Answer

The region $R$ is defined by $0 \leq x \leq 2$ and $0 \leq y \leq 2x$. To integrate first with respect to $y$, the inner integral's limits must be from $0$ to $2x$, and the outer integral's limits are from $0$ to $2$, matching option A.

Set Up ∬Rf(x,y)dA\iint _ { R } f ( x , y ) d A∬R​f(x,y)dA

Set Up $\iint _ { R } f ( x , y ) d A$