Volume of a sphere is given by V=4/3πr³ , where r is its radius . If radius changes by 1.2% , then find the change in the volume ?

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Volume of a sphere is given by V=4/3πr³ , where r is its radius . If radius changes by 1.2% , then find the change in the volume ?

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Answer 1 · 2018-06-27T14:13:48

Volume goes with the cube of linear scaling.

${1.012}^{3} = 1.036433728,$ $1.012^3=1.036433728,$ so volume increases about $3.64 % .$ $3.64%.$

Answer 2 · 2018-06-27T14:17:31

The change in volume is proportional to the original volume and approximately equal to $0.0364 \cdot Initial Volume$ $0.0364 * "Initial Volume"$ .

Explanation:

Let $V_{new}$ $V_"new"$ be the volume of the new sphere of the radius change and let $r_{new}$ $r_"new"$ be the new radius.

$r_{new} = r + 1.2 % of r = r + \frac{1.2}{100} r = r (1 + \frac{1.2}{100}) = \frac{101.2}{100} r$ $r_"new" = r+1.2%"of "r=r+1.2/100 r=r(1+1.2/100)=101.2/100r$

$V_{new} = \frac{4 π r_{new}^{3}}{3} = \frac{4 π {(\frac{101.2}{100})}^{3} r^{3}}{3} = {(\frac{101.2}{100})}^{3} \frac{4 π r^{3}}{3}$ $V_"new" = (4pir_"new"^3)/3=(4pi(101.2/100)^3r^3)/3=(101.2/100)^3 (4pir^3)/3$

$V_{new} = {(\frac{101.2}{100})}^{3} V$ $V_"new" = (101.2/100)^3 V$

$Change = V_{new} - V = V [{(\frac{101.2}{100})}^{3} - 1] \approx 0.0364 V$ $"Change" = V_"new" - V = V[(101.2/100)^3-1]~~0.0364V$

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Volume of a sphere is given by V=4/3πr³ , where r is its radius . If radius changes by 1.2% , then find the change in the volume ?

2 Answers

Explanation:

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