Question

how do you find ∫ -3 to 3 of sqrt(9-(x^2))?

Answer 1

i know you have to find the antiderivative first but like how do you do that for this?

Answer 2

\[y=\sqrt{9-x^2} \\ y^2=9-x^2 \\ x^2+y^2=9 \\ \\]


x^2+y^2=9 is a circle with center (0,0) and radius 3


y=sqrt(9-x^2) is the top part of the circle with center (0,0) and radius 3

find the area enclosed by the circle x^2+y^2=9 and half it since you have half a circle

that is you have this:

pretend I drew a lovely half circle for you