The area of the region bounded by the curve y | vedclass

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

Question

(B) The region is bounded by the curve $y = x^3$,the line $y = 8$,and the $y$-axis $(x = 0)$.To find the area with respect to the $y$-axis,we express

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The region is bounded by the curve $y = x^3$,the line $y = 8$,and the $y$-axis $(x = 0)$.To find the area with respect to the $y$-axis,we express $x$ in terms of $y$: $x = y^{1/3}$.The limits for $y$ are from $0$ to $8$.Required Area $= \int\limits_{0}^{8} x \, dy = \int\limits_{0}^{8} y^{1/3} \, dy$$= \left[ \frac{y^{1/3 + 1}}{1/3 + 1} ight]_{0}^{8} = \left[ \frac{y^{4/3}}{4/3} ight]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} ight]_{0}^{8}$$= \frac{3}{4} (8^{4/3} - 0^{4/3}) = \frac{3}{4} ((2^3)^{4/3}) = \frac{3}{4} (2^4) = \frac{3}{4} 	imes 16 = 12 	ext{ sq. units.}$

The area of the region bounded by the curve $y = x^3$,and the lines $y = 8$ and $x = 0$ is

Similar Questions

The area of one region included between the $sine$ and $cosine$ curves is given by

Let $y = g(x)$ be the inverse of a bijective mapping $f : R \rightarrow R$ defined by $f(x) = 3x^3 + 2x$. The area bounded by the graph of $g(x)$,the $x-$axis,and the ordinate at $x = 5$ is:

The area of the region bounded by the parabola $y^2 = 27x$ and the line $x = 1$ is . . . . . . sq. units. (in $\sqrt{3}$)

The area of the region bounded by the lines $y=2x+1$,$y=3x+1$ and $x=4$ is

The area bounded by the curve $y=x^{3}$,the $X$-axis and the lines $x=1$ and $x=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module