Question
Find the area enclosed by the curve $\text{x}=3\cos\text{t}, \text{y}=2\sin\text{t}.$
✓
Answer
The given curve $\text{x}=3\cos\text{t}, \text{y}=2\sin\text{t}$in t respesent the parmetric of the elipse. Eliminating the t, we get $\frac{\text{x}^{2}}{9}+\frac{\text{y}^{2}}{4}=\cos^{2}\text{t}+\sin^{2}\text{t}=1$ This represents the cartesion eqution of the elioes with center (0, 0). The coordinates of the vertics are (3, 0) and (0, 2)
Required area = Area of the shaded region = 4 × Area of the region OABO $=4\times \int\limits_{0}^{3}\text{y}\text{dx} $ $=4\times \int\limits_{0}^{3}\sqrt{4(1-\frac{\text{x}^{2}}{9})}\text{dx}$ $=4\times \frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^{2}}{})\text{dx}$ $=\frac{8}{3}\Big[(0+\frac{9}{2}\sin^{-1}1)-(0+0)\Big] $ $=\frac{8}{3}\times\frac{9}{2}\times\frac{\pi}{2}$ $=6\pi\ \text{sq.}\ \text{units}$
Required area = Area of the shaded region = 4 × Area of the region OABO $=4\times \int\limits_{0}^{3}\text{y}\text{dx} $ $=4\times \int\limits_{0}^{3}\sqrt{4(1-\frac{\text{x}^{2}}{9})}\text{dx}$ $=4\times \frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^{2}}{})\text{dx}$ $=\frac{8}{3}\Big[(0+\frac{9}{2}\sin^{-1}1)-(0+0)\Big] $ $=\frac{8}{3}\times\frac{9}{2}\times\frac{\pi}{2}$ $=6\pi\ \text{sq.}\ \text{units}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
→$\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
If $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt},$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{ y}=0.$
→Show that the matrix, $A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]$ satisfies the equation,
$A ^3- A ^2-3 A- I _3= O$. Hence, find $A ^{-1}$
→$A ^3- A ^2-3 A- I _3= O$. Hence, find $A ^{-1}$
A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area.
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$
→$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$
Find the vector and Cartesian equations of the line through the point (1, 2, – 4) and perpendicular to the two lines.
$\overrightarrow{\text{r}} = (8\hat{\text{i}} - 19\hat{\text{j}} + 10\hat{\text{k}})+\lambda(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}$ and $\overrightarrow{\text{r}} = (15\hat{\text{i}} - 29\hat{\text{j}} + 5\hat{\text{k}})+\mu(3\hat{\text{i}} - 8\hat{\text{j}} + 5\hat{\text{k})}.$
→$\overrightarrow{\text{r}} = (8\hat{\text{i}} - 19\hat{\text{j}} + 10\hat{\text{k}})+\lambda(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}$ and $\overrightarrow{\text{r}} = (15\hat{\text{i}} - 29\hat{\text{j}} + 5\hat{\text{k}})+\mu(3\hat{\text{i}} - 8\hat{\text{j}} + 5\hat{\text{k})}.$
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
→Maximum Z = x - 5y + 20
Subject to
$\text{x}-\text{y}\geq0$
$-\text{x}+2\text{y}\geq2$
$\text{x}\geq3$
$\text{y}\geq4$
$\text{x},\text{y}\geq0$
→Subject to
$\text{x}-\text{y}\geq0$
$-\text{x}+2\text{y}\geq2$
$\text{x}\geq3$
$\text{y}\geq4$
$\text{x},\text{y}\geq0$
On a multiple choice examination with three possible answers (out of which only one is correct) fo each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Mother, father and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P (A/B) and P (B/A).