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

The Correct Option is B

To find the area enclosed between the curve $x^2 = 4y$ and the line $x = y$, we begin by identifying the points of intersection. Solving both equations simultaneously: 

$x^2 = 4y$ and $x = y$

Substitute $y = x$ into $x^2 = 4y$:

$x^2 = 4x \implies x^2 - 4x = 0 \implies x(x - 4) = 0$

This gives $x = 0$ or $x = 4$. Hence, $y = 0$ or $y = 4$, giving the points (0,0) and (4,4).

Now, calculate the area between the curve and line using integration:

1. The parabola: $y = \frac{x^2}{4}$

2. Line: $y = x$

Determine the area from $x = 0$ to $x = 4$:

$\text{Area} = \int_{0}^{4} (x - \frac{x^2}{4})\, dx$

$= \int_{0}^{4} x\, dx - \int_{0}^{4} \frac{x^2}{4}\, dx$

Evaluate each integral separately:

$\int_{0}^{4} x\, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} - 0 = 8$

$\int_{0}^{4} \frac{x^2}{4}\, dx = \frac{1}{4} \left[\frac{x^3}{3}\right]_{0}^{4} = \frac{1}{4}\cdot\frac{64}{3} = \frac{16}{3}$

Hence, the area is $8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}$ square units.

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.