The area bounded by the curve x=4-y^2 and the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given curve is $x = 4 - y^2$, which is a parabola opening to the left with its vertex at $(4, 0)$.To find the intersection points with the $Y$

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq. unit}$

Vedclass · Answer

(B) The given curve is $x = 4 - y^2$, which is a parabola opening to the left with its vertex at $(4, 0)$.To find the intersection points with the $Y$-axis, we set $x = 0$:$0 = 4 - y^2 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.Thus, the curve intersects the $Y$-axis at $(0, 2)$ and $(0, -2)$.The area bounded by the curve and the $Y$-axis is symmetric about the $X$-axis.Area $= \int_{-2}^{2} x \, dy = \int_{-2}^{2} (4 - y^2) \, dy$.Due to symmetry, Area $= 2 \int_{0}^{2} (4 - y^2) \, dy$.$= 2 \left[ 4y - \frac{y^3}{3} ight]_{0}^{2}$.$= 2 \left( (4(2) - \frac{2^3}{3}) - (0 - 0) ight)$.$= 2 \left( 8 - \frac{8}{3} ight) = 2 \left( \frac{24 - 8}{3} ight) = 2 \left( \frac{16}{3} ight) = \frac{32}{3} 	ext{ sq. unit}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The area bounded by the curve $x=4-y^2$ and the $Y$-axis is

Similar Questions

The area of the region bounded by the curve $y = \tan x$,the $X$-axis,and the line $x = \frac{\pi}{3}$ is

The area of the region bounded by the curve $y=x^3$, its tangent at $(1,1)$, and the $X$-axis is

The values of a function $f(x)$ at different values of $x$ are as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$
$f(x)$ $2$ $3$ $6$ $11$ $18$ $27$

Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is:

The area bounded by the parabola $y^2=x$,the straight line $y=4$ and the $Y$-axis is

The area of the region bounded by the curve $y^2 = 4x$,the $Y$-axis,and the line $y = 3$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module