0-r-r-2-x-2-dx-

Question Number 167512 by MikeH last updated on 18/Mar/22

\int_{0}^{r} \sqrt{r^{2} - x^{2}} d x

Commented by mr W last updated on 18/Mar/22

= a r e a o f a q u a t e r c i r c l e = \frac{π r^{2}}{4}

Commented by MikeH last updated on 18/Mar/22

thanks sir . please can you

look at Q 167520

Answered by LEKOUMA last updated on 18/Mar/22

P o s o n s x = r \sin t \Rightarrow d t = r \cos t d t

s i x = 0 \Rightarrow t = 0

x = r \Rightarrow t = \frac{π}{2}

= \int_{0}^{\frac{π}{2}} \sqrt{r^{2} - r^{2} \sin^{2} t} \times r \cos t d t

= \int_{0}^{\frac{π}{2}} \sqrt{r^{2} (1 - \sin^{2} t)} \times r \cos t d t

= \int_{0}^{\frac{π}{2}} r^{2} \cos^{2} t d t = r^{2} \int_{0}^{\frac{π}{2}} \cos^{2} t d t

= r^{2} \int_{0}^{\frac{π}{2}} (\frac{1 + \cos 2 t}{2}) d t = \frac{1}{2} r^{2} (\int_{0}^{\frac{π}{2}} d t + \int_{0}^{\frac{π}{2}} \cos 2 t d t)

= \frac{1}{2} r^{2} ({[t]}_{0}^{\frac{π}{2}} + {[\frac{1}{2} \sin 2 t]}_{0}^{\frac{π}{2}})

= \frac{1}{2} r^{2} (\frac{π}{2}) = \frac{r^{2} π}{4}

Commented by MikeH last updated on 18/Mar/22

thanks a lot

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