Question

The area (in sq. units) of the region given by \( R = \left\{ (x, y) : \dfrac{y^2}{2} \leq x \leq y + 4 \right\} \) is

• A. \( 16 \)
• B. \( 18 \)
• C. \( 24 \)
• D. \( 30 \)

Hint:

Always find the bounds for both variables before setting up the double integral. Pay attention to inequalities to determine correct limits.

Answer 1

The Correct Option is B

We are given a region: \[ R = \left\{ (x, y) : \frac{y^2}{2} \leq x \leq y + 4 \right\} \] Step 1: Understand the region For a fixed \( y \), \( x \) ranges from \( \frac{y^2}{2} \) to \( y + 4 \). We must determine the limits of \( y \) where the region exists. So we find the values of \( y \) for which: \[ \frac{y^2}{2} \leq y + 4 \Rightarrow y^2 \leq 2y + 8 \Rightarrow y^2 - 2y - 8 \leq 0 \Rightarrow (y - 4)(y + 2) \leq 0 \] This gives: \[ -2 \leq y \leq 4 \] Step 2: Set up the double integral for area \[ \text{Area} = \int_{-2}^{4} \int_{\frac{y^2}{2}}^{y + 4} dx \, dy \] First integrate w.r.t \( x \): \[ = \int_{-2}^{4} \left[ x \right]_{\frac{y^2}{2}}^{y + 4} \, dy = \int_{-2}^{4} \left( y + 4 - \frac{y^2}{2} \right) dy \] Step 3: Simplify and integrate \[ = \int_{-2}^{4} \left( -\frac{y^2}{2} + y + 4 \right) dy = \int_{-2}^{4} \left( -\frac{y^2}{2} + y + 4 \right) dy \] Break it into parts: \[ = -\frac{1}{2} \int_{-2}^{4} y^2 \, dy + \int_{-2}^{4} y \, dy + \int_{-2}^{4} 4 \, dy \] Compute each: \[ \int_{-2}^{4} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-2}^{4} = \frac{64}{3} - \left( -\frac{8}{3} \right) = \frac{72}{3} = 24 \] \[ \int_{-2}^{4} y \, dy = \left[ \frac{y^2}{2} \right]_{-2}^{4} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6 \] \[ \int_{-2}^{4} 4 \, dy = 4(4 - (-2)) = 4 \times 6 = 24 \] Now substitute: \[ = -\frac{1}{2}(24) + 6 + 24 = -12 + 6 + 24 = \boxed{18} \]