Question

The area enclosed between the curve $y=1+x^2$, the y-axis and the straight line $y=5$ is given by

• A. $\frac{14}{3}\mathrm{sq}\mathrm{units}$
• B. $\frac{7}{3}\mathrm{sq}\mathrm{units}$
• C. $5\mathrm{sq}\mathrm{units}$
• D. $\frac{16}{3}\mathrm{sq}\mathrm{units}$

Answer 1

The correct option is D

$\frac{16}{3}\mathrm{sq}\mathrm{units}$

Explanation for correct option:

Determining the area under the curve:

Given the equation of the line is $y=5$

Equation of curve is $y=1+x^2...\left(1\right)$

$\therefore x=\sqrt{y-1}$

Put $x=0$ in equation $\left(1\right)$ we get,

$y=1$

Therefore the limit points are $y=1$ and $y=5$

Now the required area enclosed between the given curves is,

$\begin{array}{rcl}\mathrm{Area}& =& {\int }_1^5\mathrm{x}d\mathrm{y}\\ & =& {\int }_1^5\sqrt{\mathrm{y}-1}d\mathrm{y}\\ & =& \frac{2}{3}{\left[{\left(\mathrm{y}-1\right)}^{\frac{3}{2}}\right]}_1^5\\ & =& \frac{2}{3}\left[{\left(4\right)}^{\frac{3}{2}}-0\right]\\ & =& \frac{2}{3}\times {\left(2\right)}^3\\ & =& \frac{16}{3}\mathrm{square}\mathrm{units}\end{array}$

Therefore the area enclosed is equal to $\frac{16}{3}\mathrm{square}\mathrm{units}$.

Hence, the correct answer is option (D).