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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Hello MHB,

This is an exemple I do not understand.

if \(\displaystyle R={(x,y)|-1 \leq x \leq 1, -2 \leq y \leq 2}\), evaluate the integral

\(\displaystyle \int\int_R \sqrt{1-x^2}dA\) (It is suposed to be R at down for 'rectangle' if I understand correct.)

they solve it like this \(\displaystyle \int\int_R \sqrt{1-x^2}dA = \frac{1}{2} \pi(1)^2 * 4 = 2\pi\)

I don't understand how they do it and

The defination says:

"If \(\displaystyle f(x,y) \geq0\), then the volume V of the solid that lies above the rectangle R and below the surface \(\displaystyle z=f(x,y)\) is

\(\displaystyle V=\int\int_R f(x,y)dA\)"

What I can note is that \(\displaystyle \sqrt {1-x^2} \geq0\) and \(\displaystyle z=\sqrt{1-x^2} <=> x^2+z^2=1\) so then the volume lies above the rectangle and below \(\displaystyle z= \sqrt{1-x^2}\) but I get confused with dA, integrate respect to area?

Regards,

This is an exemple I do not understand.

if \(\displaystyle R={(x,y)|-1 \leq x \leq 1, -2 \leq y \leq 2}\), evaluate the integral

\(\displaystyle \int\int_R \sqrt{1-x^2}dA\) (It is suposed to be R at down for 'rectangle' if I understand correct.)

they solve it like this \(\displaystyle \int\int_R \sqrt{1-x^2}dA = \frac{1}{2} \pi(1)^2 * 4 = 2\pi\)

I don't understand how they do it and

The defination says:

"If \(\displaystyle f(x,y) \geq0\), then the volume V of the solid that lies above the rectangle R and below the surface \(\displaystyle z=f(x,y)\) is

\(\displaystyle V=\int\int_R f(x,y)dA\)"

What I can note is that \(\displaystyle \sqrt {1-x^2} \geq0\) and \(\displaystyle z=\sqrt{1-x^2} <=> x^2+z^2=1\) so then the volume lies above the rectangle and below \(\displaystyle z= \sqrt{1-x^2}\) but I get confused with dA, integrate respect to area?

Regards,

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