Question

find the volume of a right circular cone generated by revolving ghe line y=(3/4)x about x axis between the ordinates x=0 to x=4​

Answer 1

The volume of a right circular cone generated by revolving the line is 15π

Given :

The equation of the line y = (3/4)x

To find :

The volume of a right circular cone generated by revolving the line y = (3/4)x about x axis between the ordinates x = 0 to x = 4

Solution :

Step 1 of 2 :

Write down the given equation of the line

Here the given equation of the line is

$\displaystyle \sf{y = \frac{3x}{4} }$

Step 2 of 2 :

Find the volume of a right circular cone generated by revolving the line

$\displaystyle \sf{y = \frac{3x}{4} }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \frac{3}{4} }$

Hence the required volume of a right circular cone generated by revolving the line

$\displaystyle = \sf2 \pi\int\limits_{0}^{4} y \sqrt{1 + { \bigg( \frac{dy}{dx} \bigg)}^{2} } \, dx$

$\displaystyle = \sf2 \pi\int\limits_{0}^{4} \frac{3x}{4} \sqrt{1 + { \bigg( \frac{3}{4} \bigg)}^{2} } \, dx$

$\displaystyle = \sf2 \pi \times \frac{3}{4} \int\limits_{0}^{4} x \sqrt{1 + \frac{9}{16} } \, dx$

$\displaystyle = \sf \frac{3\pi}{2} \int\limits_{0}^{4} x \sqrt{ \frac{16 + 9}{16} } \, dx$

$\displaystyle = \sf \frac{3\pi}{2} \int\limits_{0}^{4} x \sqrt{ \frac{25}{16} } \, dx$

$\displaystyle = \sf \frac{3\pi}{2} \int\limits_{0}^{4} \frac{5x}{4} \, dx$

$\displaystyle = \sf \frac{3\pi}{2} \times \frac{5}{4} \int\limits_{0}^{4} x \, dx$

$\displaystyle = \sf \frac{15\pi}{8} \times \bigg[ \frac{ {x}^{2} }{2} \bigg]_{0}^{4}$

$\displaystyle = \sf \frac{15\pi}{8}\bigg[ \frac{ {4}^{2} }{2} - \frac{ {0}^{2} }{2} \bigg]$

$\displaystyle = \sf \frac{15\pi}{8} \times \frac{16}{2}$

$\displaystyle = \sf 15\pi$

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