Question

The area of the region bounded by the curve $y^2 = x^3$, the y-axis and the lines $y = 1$ and $y = 8$ is

• A. $\frac{155}{3}$ sq. units
• B. $\frac{93}{5}$ sq. units
• C. $93$ sq. units
• D. $155$ sq. units

Hint:

When calculating areas bounded by the y-axis and horizontal lines ($y=c$ to $y=d$), always rewrite the function as $x = f(y)$ and integrate with respect to $dy$. This avoids having to split the integral into multiple parts or dealing with more complex inverse functions if you tried integrating with respect to $dx$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the area of a region bounded by a curve and horizontal lines. Since the boundaries are defined by $y$-values ($y=1$ to $y=8$) and the y-axis ($x=0$), it is most straightforward to integrate with respect to $y$.

Step 2: Key Formula or Approach:
The area $A$ between a curve $x = f(y)$, the y-axis, and the horizontal lines $y = c$ and $y = d$ is given by the definite integral: $A = \int_{c}^{d} x \, dy$ First, express $x$ as a function of $y$ from the given curve equation. Then perform the definite integration.

Step 3: Detailed Explanation:
1. Express $x$ in terms of $y$: The equation of the curve is given as $y^2 = x^3$. We need to solve for $x$: Taking the cube root of both sides gives: \[ x = (y^2)^{1/3} = y^{2/3} \] 2. Set up the integral: The region is bounded by the y-axis ($x=0$) and the horizontal lines $y=1$ and $y=8$. Since $y$ is positive in this interval, $x = y^{2/3}$ is also positive, meaning the curve lies to the right of the y-axis. The area $A$ is: \[ A = \int_{1}^{8} x \, dy \] Substitute $x = y^{2/3}$: \[ A = \int_{1}^{8} y^{2/3} \, dy \] 3. Evaluate the integral: Use the power rule for integration: $\int y^n \, dy = \frac{y^{n+1}}{n+1}$: \[ \int y^{2/3} \, dy = \frac{y^{(2/3) + 1}}{(2/3) + 1} = \frac{y^{5/3}}{5/3} = \frac{3}{5} y^{5/3} \] Now, evaluate this from $y=1$ to $y=8$: \[ A = \left[ \frac{3}{5} y^{5/3} \right]_{1}^{8} \] \[ A = \frac{3}{5} \left( 8^{5/3} - 1^{5/3} \right) \] Let's simplify $8^{5/3}$. We know $8 = 2^3$, so: $8^{5/3} = (2^3)^{5/3} = 2^{3 \times (5/3)} = 2^5 = 32$. And $1$ to any power is $1$. Substitute these back: \[ A = \frac{3}{5} (32 - 1) \] \[ A = \frac{3}{5} (31) \] \[ A = \frac{93}{5} \]

Step 4: Final Answer:
The area of the bounded region is $\frac{93}{5}$ sq. units.