# Thread: i have no idea where to even begin with this crazy integral question about unit circl

1. ## i have no idea where to even begin with this crazy integral question about unit circl

we have already prooved that pi is the area of the unit disk and pi=2(integral from -1 to 1) square root(1-x^2) dx...use properties of the integral to compute the following in terms of pi
a. (integral from -3 to 3) square root (9-x^2) dx
b. (integral from 2 to 0) square root (1- 1/4 x^2) dx
c. (integral from -2 to 2) (x-3) square root (4-x^2) dx

im sorry about the poor notation, but i have no other idea how to write it.....i appreciate any help whatsoeever and thanks in advance.

2. Originally Posted by twostep08
we have already prooved that pi is the area of the unit disk and pi=2(integral from -1 to 1) square root(1-x^2) dx...use properties of the integral to compute the following in terms of pi
a. (integral from -3 to 3) square root (9-x^2) dx
b. (integral from 2 to 0) square root (1- 1/4 x^2) dx
c. (integral from -2 to 2) (x-3) square root (4-x^2) dx

im sorry about the poor notation, but i have no other idea how to write it.....i appreciate any help whatsoeever and thanks in advance.
You have $\pi = 2 \int_{-1}^{1} \sqrt{1 - x^2} \, dx = 4 \int_{0}^{1} \sqrt{1 - x^2} \, dx$.

a. Substitute $u = \frac{x}{3}$.

b. $\int_{2}^{0} \sqrt{1 -\frac{x^2}{4}} \, dx = - \int_{0}^{2} \sqrt{1 -\frac{x^2}{4}} \, dx$. Substitute $u = \frac{x}{2}$.

c. $\int_{-2}^{2} (x - 3) \sqrt{4 - x^2} \, dx = \int_{-2}^{2} x \sqrt{4 - x^2} \, dx - 3 \int_{-2}^{2}\sqrt{4 - x^2} \, dx = I_1 - 3 I_2$.

To calculate $I_1$ substitute $u = 4 - x^2$. To calculate $I_2$, think about b.

3. oh sorry i didnt mention before, i cant use u substitution...