Question

The area (in square units) enclosed between the curve $x^2 = 4y$ and the line $x = y$ is:

• A. 8
• B. $\frac{8}{3}$
• C. 16
• D. $\frac{16}{3}$

Answer 1

Answer: (B)

$\frac{8}{3}$

Answer 2

We are tasked with finding the area enclosed between the parabola $x^2 = 4y$ and the line $x = y$.

Find the points of intersection: The curve $x^2 = 4y$ can be written as $y = \frac{x^2}{4}$. Setting $y = x$ (for the line), we get:

$x = \frac{x^2}{4} \Rightarrow 4x = x^2 \Rightarrow x(x - 4) = 0.$ Thus, $x = 0$ and $x = 4$. The region is enclosed between $x = 0$ and $x = 4$.

Setup the area integral: The area is computed as the integral of the difference between the top curve $(y = x)$ and the bottom curve $(y = \frac{x^2}{4})$

$Area = \int_0^4 \left( x - \frac{x^2}{4} \right) dx.$

Compute the integral:

$\int_0^4 \left( x - \frac{x^2}{4} \right) dx = \int_0^4 x dx - \int_0^4 \frac{x^2}{4} dx.$

Compute each term separately:

$\int_0^4 x dx = \left[ \frac{x^2}{2} \right]_0^4 = \frac{4^2}{2} - 0 = \frac{16}{2} = 8.$

$\int_0^4 \frac{x^2}{4} dx = \frac{1}{4} \int_0^4 x^2 dx = \frac{1}{4} \left[ \frac{x^3}{3} \right]_0^4 = \frac{1}{4} \left( \frac{4^3}{3} - 0 \right) = \frac{1}{4} \times \frac{64}{3} = \frac{16}{3}.$

Now subtract the results:

$Area = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.$

Thus, the enclosed area is $\frac{8}{3}$ square units.