To solve the problem, we need to find the area of the shaded region bounded by the curve \( y = x^2 \), the vertical line \( x = 2 \), the y-axis \( x = 0 \), and the region between \( y = 0 \) and \( y = 4 \).
1. Understand the Geometry:
The curve given is \( y = x^2 \). To express the integral in terms of \( y \), we need to solve for \( x \):
From \( y = x^2 \), we get \( x = \sqrt{y} \)
The vertical line \( x = 2 \) intersects the parabola at \( y = 4 \), since \( y = (2)^2 = 4 \)
So, the bounds for \( y \) go from 0 to 4.
2. Area with Respect to y:
The shaded region is bounded on the left by the parabola \( x = \sqrt{y} \) and on the right by the vertical line \( x = 2 \). So the horizontal strip from left to right spans from \( x = \sqrt{y} \) to \( x = 2 \).
Thus, the width of the strip is \( (2 - \sqrt{y}) \), and height \( dy \).
So the area is:
\( A = \int_{0}^{4} (2 - \sqrt{y}) \, dy \)
But this is not one of the listed options. However, the key idea is to represent the shaded area in terms of the correct integral expression format.
3. Compare with Options:
Option (D): \( \int_0^4 \sqrt{y} \, dy \) — This would represent the area under the curve \( x = \sqrt{y} \), not the full strip up to \( x = 2 \).
Option (B): \( \int_0^2 \sqrt{y} \, dy \) — Upper limit should be 4, not 2.
Options (A) and (C): \( \int_0^2 x^2 \, dx \) — This represents the area under the parabola \( y = x^2 \) from \( x = 0 \) to \( x = 2 \), which matches our shaded region!
4. Conclusion:
The area of the shaded region is correctly given by:
\( \int_0^2 x^2 \, dx \)
Final Answer:
The correct option is (A) \( \int_0^2 x^2 \, dx \).
