Calculating the Area of a Circle in the First Quadrant | Integration Homework

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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?
 
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gtfitzpatrick said:

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.
 
Dick said:
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":confused:
 
gtfitzpatrick said:
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":confused:

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.
 
thanks a million Dick. :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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