What is area y=sqrt(9-x^2),-1<= x<= 1 ?

The answer to area y=sqrt(9-x^2),-1<= x<= 1 is 9arcsin(1/3)+2sqrt(2)

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area y=sqrt(9-x^2),-1<= x<= 1

area

y = 9 - x^{2}, - 1 \leq x \leq 1

9 arcsin (\frac{1}{3}) + 22

Decimal

5.88695 \dots

Solution steps

Apply the area formula:

\int_{- 1}^{1} 9 - x^{2} d x

Solve

\int_{- 1}^{1} 9 - x^{2} d x : 9 arcsin (\frac{1}{3}) + 22

The area is:

= 9 arcsin (\frac{1}{3}) + 22

\int 2 x^{2} sin (\frac{x}{3}) d x

\frac{d}{d x} (9 x - 18)

\int sin (4 x) ln (sin (4 x)) d x

\int (3 t + t^{2}) sin (2 t) d t

\frac{d}{d x} (6 x x^{3} + 3 x)

What is area y=sqrt(9-x^2),-1<= x<= 1 ?
The answer to area y=sqrt(9-x^2),-1<= x<= 1 is 9arcsin(1/3)+2sqrt(2)