Question

The area of the region bounded by y² = 8x and y² = 16(3 – x) is equal to

• A. \(\frac{32}{3}\)
• B. \(\frac{40}{3}\)
• C. 16
• D. 19

Answer 1

The Correct Option is C

 To find the area of the region bounded by the curves \(y^2 = 8x\) and \(y^2 = 16(3 - x)\), we need to identify the points of intersection and then integrate to find the area between the curves.

Find points of intersection:

Equate \(y^2 = 8x\) and \(y^2 = 16(3 - x)\):

1. \(8x = 16(3 - x)\)

This simplifies to:

1. \(8x = 48 - 16x\)

Rearranging gives:

1. \(24x = 48 \quad \Rightarrow \quad x = 2\)

Find potential bounds on the x-axis:

From the above equation, we know that when \(x = 2\), both curves have the same \(y^2\) value. Now, substitute \(x = 0\) to check:

1. \(y^2 = 8 \times 0 = 0 \Rightarrow y = 0\)
   \(y^2 = 16(3 - 0) = 48\)

The curves intersect at \(x = 2\) and \(y = \pm \sqrt{16} = \pm 4\).

Calculate the area between the curves:

The bounds of integration are \(x = 0\) to \(x = 3\). The curve \(y^2 = 8x\) is 'inside' from \(x = 0\) to \(x = 2\), and 'outside' beyond \(x = 2\). Integrate the difference to find the area:

\[\text{Area} = 2 \int_{0}^{2} \left(\sqrt{16(3 - x)} - \sqrt{8x}\right) \, dx\]

Let's evaluate the integral:

• For \(y^2 = 16(3 - x)\):
• For \(y^2 = 8x\):

Conclusion:

The area of the region bounded by the given curves is 16.

Thus, the correct answer is: 16