bounded-by-the-curve-y-4-x-y-0-y-1-

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

b o u n d e d b y t h e c u r v e y = \sqrt{4 - x} y = 0 y = 1

Commented by jagoll last updated on 28/Feb/20

Area = \int_{0}^{1} (4 - y^{2}) dy

= 4 y - \frac{y^{3}}{3} ∣_{0}^{1} = \frac{11}{3}

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

c o r r e c t i s 3 + \frac{2}{3} = \frac{11}{3}

Commented by jagoll last updated on 28/Feb/20

this question request volume

or area ?

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

a r e a

Commented by jagoll last updated on 28/Feb/20

Commented by jagoll last updated on 28/Feb/20

where this area ?

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

f r o m y = 0 t o y = 1 .

n o t y o u r m e t h o d i s w r o n g, b u t y o u r

c a l c u l a t i o n i s w r o n g . t h e r e s u l t i s

n o t \frac{8}{3}, b u t \frac{11}{3} .

\int (4 - y^{2}) d y = 4 y - \frac{y^{3}}{3} \neq 4 y - \frac{4 y^{3}}{3}

b u t w e c a n a l s o g e t t h e r e s u l t w i t h o u t

i n t e g r a l, t h e a r e a i s 3 + \frac{2}{3} = \frac{11}{3} .

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

Commented by jagoll last updated on 28/Feb/20

hahaha \dots yes sir . thank 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 = \sqrt{4 - x}, y = 0, y = 1, x = 0

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