- #1

- 244

- 0

[tex] y=\sqrt{a^2-x^2} [/tex] where -a<=x<=a is a semicircle with r=a

show why [tex]\int_{-a}^{a} \sqrt{a^2-x^2}\,dx = \frac{1}{2} \pi a^2 [/tex]

(I think that's how its written)

- Thread starter tandoorichicken
- Start date

- #1

- 244

- 0

[tex] y=\sqrt{a^2-x^2} [/tex] where -a<=x<=a is a semicircle with r=a

show why [tex]\int_{-a}^{a} \sqrt{a^2-x^2}\,dx = \frac{1}{2} \pi a^2 [/tex]

(I think that's how its written)

- #2

- 362

- 0

What's the area of a half circle?

- #3

- 650

- 1

Do you know the area of circle

if yes then u can find the area of semicircle

or if u dont know n u want to calulate the integral

substitute

x=asinz

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