Solution: Area bounded by the curve, $y^{2}=4 x, y$-axis, and $y=3$ is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{3} x d y \\ &=\int_{0}^{3}...
Choose the correct answer: Area of the region bounded by the curve $y^{2}=4x$, y-axis and the line $y=3$ is
Choose the correct answer: Area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x=0$ and $x=2$ is
A.$\pi$
B.$\frac\pi2$
C.$\frac\pi3$
D.$\frac\pi4$
Solution: The area bounded by the circle and the lines, $x=0$ and $x=2$, in the first quadrant is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{2} y d x \\...
Find the area of the region bounded by the curve $y^{2}=4x$
Solution: The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$. Area of OACO is symmetrical about $x$-axis. $\therefore$ Area of $O A C O=2$ (Area of...
Find the area bounded by the curve $x=4y$ and the line $x=4y-2$
Solution: The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area OBAO. Let's say $A$ and $B$ be the points of intersection of the line and parabola....
Find the area of the region bounded by the parabola $y=x^ {2}$ and the $y=\mid {x}\mid$.
Solution: The area bounded by the parabola, $x^{2}=y$, and the line, $y=|x|$, can be represented as The given area is symmetrical about $y$-axis. $\therefore$ Area $\mathrm{OACO}=$ Area...
The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$, find the value of a.
Solution: The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts. $\therefore$ Area $O A D=$ Area $A B C D$ It is observed that the given area is symmetrical about...
Find the area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cutt off by the line $x=\frac{a}{\sqrt{2}}$
Solution: The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cutt off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the ABCDA It can be observed that the area $A B C D$ is...
Find the area of the region in the first quadrant enclosed by x-axis, line $x=\sqrt{3}y$ and the circle $x^{2}+y^{2}=4$
Solution: The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is...
Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
Solution: The given eq. of the ellipse can be represented as $\begin{array}{l} \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \\ \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}} \end{array}$ It can be observed that the...
Find the area of the region bounded by the ellipse $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$
Solution: The given eq. of the ellipse, $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$ can be reprsented by It is observed that the ellipse is symmetrical about x-axis and y-axis. $\therefore$ Area...
Find the area of the region bounded by $x^{2}=4y$, $y=2$, $y=4$ and the y-axis in the first quadrant.
Solution: The area of the region bounded by $x^{2}=4y$, $y=2$, $y=4$ and the y-axis is the area of ABCD
Find the area of the region bounded by $y^{2}=9x$, $x=2$, $x=4$ and the x-axis in the first quadrant.
Solution: Area of the region bounded by the curve $y^{2}=9x$, $x=2$, $x=4$, \and the x-axis is the area of ABCD
Find the area of the region bounded by the curve $y^{2}=x$ and the lines x=1, x=4 and the x-axis in the first quadrant.
Solution: