Consider the double integral ${\int }_0^2{\int }_{y^2}^{4}f(x,y)dxdy$. Which of the following correctly expresses the integral with the order of integration reversed?

• A. ${\int }_0^2{\int }_4^{y^2}f(x,y)dxdy$
• B. ${\int }_0^4{\int }_0^{\sqrt{x}}f(x,y)dydx$
• C. ${\int }_0^2{\int }_0^{y^2}f(x,y)dxdy$
• D. ${\int }_0^4{\int }_x^{2}f(x,y)dydx$

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the problem. The goal is to reverse the order of integration for the given double integral \( \int_0^2 \int_{y^2}^{4} f(x, y) \, dx \, dy \). This involves analyzing the region of integration and rewriting the limits accordingly.
2. Step 2: Visualize the region of integration. The inner integral \( \int_{y^2}^{4} \) indicates that \( x \) ranges from \( y^2 \) to \( 4 \), while the outer integral \( \int_0^2 \) shows that \( y \) ranges from \( 0 \) to \( 2 \). Plot these bounds on the \( xy \)-plane to understand the region.
3. Step 3: Determine the new bounds for \( x \) and \( y \) when reversing the order of integration. From the visualization, \( x \) ranges from \( 0 \) to \( 4 \), and for a fixed \( x \), \( y \) ranges from \( 0 \) to \( \sqrt{x} \). This is because \( y^2 = x \) implies \( y = \sqrt{x} \).
4. Step 4: Rewrite the integral with the reversed order of integration. The new integral becomes \( \int_0^4 \int_0^{\sqrt{x}} f(x, y) \, dy \, dx \), where \( x \) is integrated first and \( y \) second.
5. Step 5: Verify the correctness of the new integral. Ensure that the new bounds correctly describe the same region of integration as the original integral. The region described by \( x \) from \( 0 \) to \( 4 \) and \( y \) from \( 0 \) to \( \sqrt{x} \) matches the original region, confirming the reversal is accurate.