The volume of a cylinder is given by V=πr^2h . If each of the radius and height of the cylinder increases by 2%, what is the increase in its volume and its surface area?

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The volume of a cylinder is given by V=πr^2h . If each of the radius and height of the cylinder increases by 2%, what is the increase in its volume and its surface area?

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Answer 1 · 2017-11-30T12:14:45

Increase in volume is $0.06 \cdot π r^{2} h (6 %)$ $0.06*pir^2h (6%)$ and increase in surface area is $0.04 \cdot 2 π r (r + h) (4 %)$ $0.04*2pi r(r+h)(4%)$

Explanation:

Let radius and height of original cylinder be $r and h$ $r and h$ respectively.

$2 %$ $2%$ increase means new radius and height will be $1.02 r, 1.02 h$ $1.02r,1.02h$

respectively. Volume of original cylinder is $V_{1} = π r^{2} h$ $V_1= pi r^2h$ and

volume of larger cylinder is $V_{2} = π {(1.02 r)}^{2} \cdot 1.02 h$ $V_2= pi(1.02r)^2*1.02h$

Increase in volume is $V_{2} - V_{1} = π {(1.02 r)}^{2} \cdot 1.02 h - π r^{2} h$ $V_2-V_1= pi(1.02r)^2*1.02h-pi r^2h$ or

$V_{2} - V_{1} = π r^{2} h ({1.02}^{3} - 1) \approx π r^{2} h \cdot 0.06 (2 d p)$ $V_2-V_1= pi r^2h(1.02^3-1)~~pir^2h*0.06(2dp)$

cubic.unit. Surface area of original cylinder is $S_{1} = 2 π r (r + h)$ $S_1= 2 pi r(r+h)$ and

surface area of larger cylinder is $S_{2} = 2 π \cdot 1.02 r (1.02 r + 1.02 h)$ $S_2= 2 pi*1.02r(1.02r+1.02h)$

$= 2 π \cdot {1.02}^{2} \cdot r (r + h)$ $=2 pi *1.02^2*r(r+h)$ .Increase in surface area is

$S_{2} - S_{1} = 2 π r (r + h) ({1.02}^{2} - 1) = 2 π r (r + h) \cdot 0.04$ $S_2-S_1= 2 pi r(r+h)(1.02^2-1)=2pi r(r+h)*0.04$ .

Increase in volume is $0.06 \cdot π r^{2} h (6 %)$ $0.06*pir^2h (6%)$

and increase in surface area is $0.04 \cdot 2 π r (r + h) (4 %)$ $0.04*2pi r(r+h)(4%)$ [Ans]

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The volume of a cylinder is given by V=πr^2h . If each of the radius and height of the cylinder increases by 2%, what is the increase in its volume and its surface area?

1 Answer

Explanation:

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