Reverse the order of integration for the following integral. $\int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { \sqrt { 8 y } } f ( x , y ) c ^ { 2 } x d y$
A) $\int _ { 0 } ^ { 2 } \int _ { x ^ { 2 } / 8 } ^ { \sqrt { x } } f ( x , y ) d y d x$
B) $\int _ { 0 } ^ { 4 } \int _ { x ^ { 2 } / 8 } ^ { \sqrt { x } } f ( x , y ) d x d y$
C) $\int _ { 0 } ^ { 4 } \int _ { x ^ { 2 } / 8 } ^ { x } f ( x , y ) d y d x$
D) $\int _ { 0 } ^ { 4 } \int _ { 8 x ^ { 2 } } ^ { \sqrt { x } } f ( x , y ) d y d x$
E) $\int _ { 0 } ^ { 4 } \int _ { x ^ { 2 } / 8 } ^ { \sqrt { x } } f ( x , y ) d y d x$

Question

Quizplus · Accepted Answer

To reverse the order of integration, we first identify the bounds for $x$ and $y$ in the original integral. The original bounds are $0 \leq y \leq 2$, $y^2 \leq x \leq \sqrt{8y}$. To reverse the order, we solve for $y$ in terms of $x$, which gives us $x = y^2$ (or $y = \sqrt{x}$) and $x = \sqrt{8y}$ (or $y = x^2/8$). The new bounds for $x$ are from the intersection points of $y^2$ and $\sqrt{8y}$, which are $0$ and $4$. Thus, the correct reversed integral is $\int _ { 0 } ^ { 4 } \int _ { x ^ { 2 } / 8 } ^ { \sqrt { x } } f ( x , y ) d y d x$.

Reverse the Order of Integration for the Following Integral ∫02∫y28yf(x,y)c2xdy\int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { \sqrt { 8 y } } f ( x , y ) c ^ { 2 } x d y∫02​∫y28y​​f(x,y)c2xdy

Reverse the Order of Integration for the Following Integral $\int _ { 0 } ^ { 2 } \int _ { y ^ { 2 } } ^ { \sqrt { 8 y } } f ( x , y ) c ^ { 2 } x d y$