Area under the curve y = √(9−x²) from 0 to 3

[gtfitzpatrick]

Homework Statement

-8$\int^{3}_{0}\sqrt{9-x^2}dx$

Homework Equations

The Attempt at a Solution

am i right in thinking this the area of a circle in the first quadrant so my answer is-8($\frac{9\pi}{4})$ = -18$\pi$

Thanks for reading?

 

[Dick]

Homework Statement

-8$\int^{3}_{0}\sqrt{9-x^2}dx$

Homework Equations

The Attempt at a Solution

am i right in thinking this the area of a circle in the first quadrant so my answer is-8($\frac{9\pi}{4})$ = -18$\pi$

Thanks for reading?

Sure it is.

 

[gtfitzpatrick]

Sure it is.

Hi Dick.
thanks for reply but...
are you saying "it sure is" or are you asking "are you sure it is?"

"sure it is" is the way my friends would sarcastically say "your wrong"

 

[Dick]

Hi Dick.
thanks for reply but...
are you saying "it sure is" or are you asking "are you sure it is?"

"sure it is" is the way my friends would sarcastically say "your wrong"

It's the not sarcastic 'it sure is'. The integral is -18*pi and you can deduce that from its being a quarter circle. You could also do it with a trig substitution and get the same thing.

 

[gtfitzpatrick]

thanks a million Dick.