The volume of a spherical cap of height h cut | vedclass

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

Question

(A) The required volume of the spherical cap is generated by revolving the area $ABCA$ of the circle $x^2 + y^2 = a^2$ about the $x$-axis.Here,$CA = h

Vedclass · Accepted Answer

$\frac{\pi}{3}h^2(3a - h)$

Vedclass · Answer

(A) The required volume of the spherical cap is generated by revolving the area $ABCA$ of the circle $x^2 + y^2 = a^2$ about the $x$-axis.Here,$CA = h$ and $OA = a$.Therefore,$OC = OA - CA = a - h$.Thus,$x$ varies from $a - h$ to $a$.The required volume $V = \int_{a - h}^{a} \pi y^2 dx$.Since $x^2 + y^2 = a^2$,we have $y^2 = a^2 - x^2$.$V = \pi \int_{a - h}^{a} (a^2 - x^2) dx$$V = \pi \left[ a^2x - \frac{x^3}{3} \right]_{a - h}^{a}$$V = \pi \left[ (a^3 - \frac{a^3}{3}) - (a^2(a - h) - \frac{(a - h)^3}{3}) \right]$$V = \pi \left[ \frac{2a^3}{3} - (a^3 - a^2h - \frac{a^3 - 3a^2h + 3ah^2 - h^3}{3}) \right]$$V = \pi \left[ \frac{2a^3}{3} - (a^3 - a^2h - \frac{a^3}{3} + a^2h - ah^2 + \frac{h^3}{3}) \right]$$V = \pi \left[ \frac{2a^3}{3} - (\frac{2a^3}{3} - ah^2 + \frac{h^3}{3}) \right]$$V = \pi (ah^2 - \frac{h^3}{3}) = \frac{\pi h^2}{3}(3a - h)$.

The volume of a spherical cap of height $h$ cut off from a sphere of radius $a$ is equal to

Similar Questions

The area of the region (in square units) bounded by the curve ${x^2} = 4y$,the line $x = 2$,and the $x$-axis is:

The area enclosed between the curve $y = \log_e(x + e)$ and the coordinate axes is

The area bounded by the circle $x^2 + y^2 = 4$,the line $x = \sqrt{3}y$,and the $x$-axis lying in the first quadrant is:

The area of the region $A = \{ (x,y) : 0 \le y \le x|x| + 1, -1 \le x \le 1 \}$ in square units is:

The area of the region under the curve $y=|\sin x-\cos x|$,$0 \leq x \leq \frac{\pi}{2}$ and above the $x$-axis,is (in square units)

Vedclass Test Series

Exam Paper Generator

Online Exam Module