## mr570 Group Title How do I find the integral from -4 to 4 of (3*sqrt(16-x^2)^2 dx ? I think you have to use u-substitution but I don't entirely understand how that works in this case. one year ago one year ago

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1. satellite73 Group Title

use nothing

2. Umangiasd Group Title

well the sqrt gets eliminated with the power 2, so...Its a simple integral

3. satellite73 Group Title

$y=\sqrt{16-x^2}$ is the upper half of a circle centered at the origin with radius 4 so $6\int_{-4}^4\sqrt{16-x^2}dx$ is 6 times the area of the upper half of a circle with radius 4

4. satellite73 Group Title

oh sorry i didn't see the square at the end if it is really the square root squared, then it is easy enough, although a rather strange way to write it

5. satellite73 Group Title

it is really $3\int_{-4}^4\sqrt{16-x^2}^2dx$?

6. Dido525 Group Title

Assuming it's what sattelite said then this becomes very simple.

7. Dido525 Group Title

|dw:1354850130324:dw| Then this become easy to integrate.