Math Help - area of a circle (integral)

1. area of a circle (integral)

use a double integral and trigonometric substitution to find the area of a circle with radius r. use the formula int(cos^2u)du = (1/2)u + (1/4)sin2u + C

so ive gotten this far(split the integral into a quarter, then will multiply by 4)...

4*int(0 to r)*int(0 to sqr[r^2-x^2]) 1 dydx

integrating sqr[r^2-x^2] seems to get messy, so i am assuming my substitution comes in here. but im not sure what it would be
would it be something like cos[sqr(r^2-x^2)/r] ?

there are also 3 other questions, which want me to find the volume of a sphere using triple integrals, and hypervolume of a hypersphere using quadruple integrals, and the volume of a hypersphere using n integrals. however, i think before i start those, i need to figure out the first basic one, but i cant.

2. For the first one you have $\int_0^r \sqrt{r^2-x^2}dx$ right? How about if you let $x=r\sin(\theta)$. Can you show this equates to $r^2\int_0^{\pi/2} \cos^2(\theta)d\theta$. You can do that right?

Yea well, I guess you got $\int_0^{r}\int_0^{\sqrt{r^2-x^2}} dydx$. Same dif right?

3. same question

Originally Posted by chrisc
use a double integral and trigonometric substitution to find the area of a circle with radius r. use the formula int(cos^2u)du = (1/2)u + (1/4)sin2u + C

so ive gotten this far(split the integral into a quarter, then will multiply by 4)...

4*int(0 to r)*int(0 to sqr[r^2-x^2]) 1 dydx

integrating sqr[r^2-x^2] seems to get messy, so i am assuming my substitution comes in here. but im not sure what it would be
would it be something like cos[sqr(r^2-x^2)/r] ?

there are also 3 other questions, which want me to find the volume of a sphere using triple integrals, and hypervolume of a hypersphere using quadruple integrals, and the volume of a hypersphere using n integrals. however, i think before i start those, i need to figure out the first basic one, but i cant.

I have the same question as you did, but I dont know how to get the rest. Did you figure it out?