Question

The area included between the parabolas y² = 4x and x² = 4y is (in square units)

A. 4/3

B. 1/3

C. 16/3

D. 8/3

Answer 1

Correct answer is C.

The blue area is what we need to compute. To do that we need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.

This means that at points of intersection

y² = 4x

\(\Rightarrow \left(\frac{x^2}{4} \right)^2 = 4x\) (from the other equation)

Let us solve this.

\(\left(\frac{x^2}{4} \right)^2 = 4x\)

x⁴ = 64x

⇒ x(x³ – 64) = 0

⇒ x = 0 or x³ = 64, i.e, x = 4

So the points of intersection are (0, 0) and (4, 4)

Now, let’s compute the area.

If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x = 4.

i.e.,