Understanding the Curves
To find the area bounded by the curves \( y^2 = 20x \) and \( x^2 = 16y \), we first rewrite these equations in more familiar forms:
- The first equation, \( y^2 = 20x \), represents a rightward-opening parabola.
- The second equation, \( x^2 = 16y \), represents an upward-opening parabola.
Finding Points of Intersection
To find the area between these curves, we need to determine the points where they intersect.
1. From \( y^2 = 20x \), we can express \( x \) in terms of \( y \):
\[
x = \frac{y^2}{20}
\]
2. Substitute this into the second equation \( x^2 = 16y \):
\[
\left(\frac{y^2}{20}\right)^2 = 16y
\]
Simplifying leads to:
\[
\frac{y^4}{400} = 16y
\]
\[
y^4 - 640y = 0
\]
\[
y(y^3 - 640) = 0
\]
Thus, \( y = 0 \) or \( y = 8\sqrt{10} \).
Finding Corresponding x-values
Using \( y = 8\sqrt{10} \) in \( x = \frac{y^2}{20} \):
\[
x = \frac{(8\sqrt{10})^2}{20} = \frac{640}{20} = 32
\]
The points of intersection are \( (0, 0) \) and \( (32, 8\sqrt{10}) \).
Calculating the Area
To find the area between the curves, we integrate the difference of the functions:
1. From \( y^2 = 20x \), we have \( y = \sqrt{20x} \).
2. From \( x^2 = 16y \), we have \( y = \frac{x^2}{16} \).
The area \( A \) is given by:
\[
A = \int_0^{32} \left( \sqrt{20x} - \frac{x^2}{16} \right) dx
\]
Calculating this integral:
1. The integral of \( \sqrt{20x} \) gives \( \frac{2}{3}(20x)^{3/2} \), evaluated from 0 to 32.
2. The integral of \( \frac{x^2}{16} \) gives \( \frac{x^3}{48} \), evaluated from 0 to 32.
After evaluation, the area calculates to \( \frac{320}{3} \) square units.
Conclusion
Thus, the area bounded by the curves is \( \frac{320}{3} \) square units, confirming option C as the correct answer.