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Question Number 83110 by 09658867628 last updated on 28/Feb/20

bounded by the curve y=(√(4-x)) y=0 y=1

$${bounded}\:{by}\:{the}\:{curve}\:{y}=\sqrt{\mathrm{4}-{x}}\:{y}=\mathrm{0}\:{y}=\mathrm{1} \\ $$

Commented by jagoll last updated on 28/Feb/20

Area = ∫_0 ^1 (4−y^2 ) dy = 4y − (y^3 /3) ∣_0 ^1 = ((11)/3)

$$\mathrm{Area}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\: \\ $$$$=\:\mathrm{4y}\:−\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{3}}\:\mid_{\mathrm{0}} ^{\mathrm{1}} =\:\frac{\mathrm{11}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 28/Feb/20

correct is 3+(2/3)=((11)/3)

$${correct}\:{is}\:\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{3}} \\ $$

Commented by jagoll last updated on 28/Feb/20

this question request volume or area?

$$\mathrm{this}\:\mathrm{question}\:\mathrm{request}\:\mathrm{volume}\: \\ $$$$\mathrm{or}\:\mathrm{area}? \\ $$

Commented by mr W last updated on 28/Feb/20

area

$${area} \\ $$

Commented by jagoll last updated on 28/Feb/20

Commented by jagoll last updated on 28/Feb/20

where this area ?

$$\mathrm{where}\:\mathrm{this}\:\mathrm{area}\:? \\ $$

Commented by mr W last updated on 28/Feb/20

from y=0 to y=1. not your method is wrong, but your calculation is wrong. the result is not (8/3), but ((11)/3). ∫(4−y^2 )dy=4y−(y^3 /3)≠4y−((4y^3 )/3) but we can also get the result without integral, the area is 3+(2/3)=((11)/3).

$${from}\:{y}=\mathrm{0}\:{to}\:{y}=\mathrm{1}. \\ $$$${not}\:{your}\:{method}\:{is}\:{wrong},\:{but}\:{your} \\ $$$${calculation}\:{is}\:{wrong}.\:{the}\:{result}\:{is} \\ $$$${not}\:\frac{\mathrm{8}}{\mathrm{3}},\:{but}\:\frac{\mathrm{11}}{\mathrm{3}}. \\ $$$$\int\left(\mathrm{4}−{y}^{\mathrm{2}} \right){dy}=\mathrm{4}{y}−\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\neq\mathrm{4}{y}−\frac{\mathrm{4}{y}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$ \\ $$$${but}\:{we}\:{can}\:{also}\:{get}\:{the}\:{result}\:{without} \\ $$$${integral},\:{the}\:{area}\:{is}\:\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{3}}. \\ $$

Commented by mr W last updated on 28/Feb/20

Commented by jagoll last updated on 28/Feb/20

hahaha...yes sir. thank you

$$\mathrm{hahaha}...\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by jagoll last updated on 28/Feb/20

but sir. i think this question not complet. my understand the question must be the area bounded the curve y = (√(4−x)) , y=0 , y=1 , x = 0

$$\mathrm{but}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{question}\:\mathrm{not}\: \\ $$$$\mathrm{complet}.\:\mathrm{my}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{the}\: \\ $$$$\mathrm{curve}\:\mathrm{y}\:=\:\sqrt{\mathrm{4}−\mathrm{x}}\:,\:\mathrm{y}=\mathrm{0}\:,\:\mathrm{y}=\mathrm{1}\:,\:\mathrm{x}\:=\:\mathrm{0} \\ $$

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