Question

If the area bounded by the curve \(x^2 = by\) and the lines \(y = 1,y = 4\) in the first quadrant is \(28\) sq. units, then the value of b is...

• A. \(36\)
• B. \(6\)
• C. \(9\)
• D. \(3\)

Hint:

Integrate x = sqrt(b) sqrt(y) from y = 1 to y = 4.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In the first quadrant, the curve \(x^2=by\) gives \(x=\sqrt{b}\sqrt{y}\). The area between the curve and the \(y\)-axis, from \(y=1\) to \(y=4\), is found by integrating \(x\) with respect to \(y\).

Step 2: Set up:
\[ A=\int_1^4 x\,dy=\sqrt b\int_1^4 y^{1/2}dy \]

Step 3: Integrate:
\[ A=\sqrt b\left[\frac23y^{3/2}\right]_1^4=\sqrt b\cdot\frac23(8-1)=\frac{14\sqrt b}{3} \]

Step 4: Use the given area:
\(\dfrac{14\sqrt b}{3}=28\), so \(\sqrt b=6\) and \(b=36\).

Step 5: Choose:
Option (A). The value \(6\) in option (B) is \(\sqrt b\), not \(b\).

Final Answer:
b = 36. \[ \boxed{36} \]