The volume of the solid formed by rotating th | vedclass

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

Question

(A) The volume $V$ of a solid generated by rotating a curve $y = f(x)$ about the $x$-axis between $x = a$ and $x = b$ is given by the formula: $V = \i

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) The volume $V$ of a solid generated by rotating a curve $y = f(x)$ about the $x$-axis between $x = a$ and $x = b$ is given by the formula: $V = \int_{a}^{b} \pi y^{2} dx$. Given the curve $y^{2} = 4x$ and the limits $x = 4$ to $x = 5$. Substituting the values into the formula: $V = \int_{4}^{5} \pi (4x) dx$ $V = 4\pi \int_{4}^{5} x dx$ $V = 4\pi \left[ \frac{x^{2}}{2} \right]_{4}^{5}$ $V = 2\pi [x^{2}]_{4}^{5}$ $V = 2\pi (5^{2} - 4^{2})$ $V = 2\pi (25 - 16)$ $V = 2\pi (9) = 18\pi$ cubic units.

The volume of the solid formed by rotating the area enclosed between the curve $y^{2}=4x$, $x=4$, and $x=5$ about the $x$-axis is (in cubic units): (in $\pi$)

Similar Questions

The area of the region bounded by the parabola $y^2 = 4x$ and its latus rectum is . . . . . . sq. units.

The area of the region $\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}$ is

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

The area (in sq. units) of the region bounded by the $X$-axis and the curve $y=1-x-6x^2$ is

If the area of the region $\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$ is $(\log_e 2) - \frac{1}{7}$,then the value of $7a - 3$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module