Question

Find the area of the region bounded by the parabola \( y^2 = 25x \) and the line \( x=5 \).

Hint:

For regions bounded by curves that are functions of \(x\), use vertical strips (\(dx\)). The area is \( \int_a^b (y_{upper} - y_{lower}) \, dx \). Utilizing symmetry can often simplify the calculation.

Answer 1

Step 1: Visualize the region.
The parabola \( y^2 = 25x \) opens to the right with its vertex at the origin. The line \( x=5 \) is a vertical line. The region is bounded on the left by the y-axis (\(x=0\)) and on the right by \(x=5\). The region is symmetric about the x-axis.
Step 2: Set up the integral.
We can find the area of the top half and double it. From \( y^2 = 25x \), the upper half is \( y = \sqrt{25x} = 5\sqrt{x} \). The limits of integration are from \( x=0 \) to \( x=5 \). \[ \text{Area} = 2 \int_0^5 5\sqrt{x} \, dx = 10 \int_0^5 x^{1/2} \, dx \] Step 3: Evaluate the integral.
Using the power rule for integration: \[ \text{Area} = 10 \left[ \frac{x^{3/2}}{3/2} \right]_0^5 = 10 \left[ \frac{2}{3}x^{3/2} \right]_0^5 \] \[ = \frac{20}{3} [5^{3/2} - 0^{3/2}] = \frac{20}{3} (5\sqrt{5}) = \frac{100\sqrt{5}}{3} \] The area is \( \frac{100\sqrt{5}}{3} \) square units.