Mass using Simpson's rule and Semi-Circle

[gucciryan]

At first, I thought it was just the integral of the area of the semi-circle (8pi) times the Simpson's rule ((1/3)(2.1+4(2.1)+2(2.3)+4(2)+2.6)dx from 0 to 4 but that wasn't the case. How do I use Simpson's Rule to find the total mass?

 

[MarkFL]

The mass \(dm\) of an element of the plate (vertical strip) will be the product of the mass density \(\rho(x)\) and the area \(A(x)\) of the strip:

[MATH]A(x)=2\sqrt{4^2-x^2}\,dx[/MATH]
[MATH]dm=2\sqrt{4^2-x^2}\rho(x)\,dx[/MATH]
Hence:

[MATH]m=2\int_0^4 \sqrt{4^2-x^2}\rho(x)\,dx[/MATH]
Can you proceed?

 

[gucciryan]

The mass \(dm\) of an element of the plate (vertical strip) will be the product of the mass density \(\rho(x)\) and the area \(A(x)\) of the strip:

[MATH]A(x)=2\sqrt{4^2-x^2}\,dx[/MATH]
[MATH]dm=2\sqrt{4^2-x^2}\rho(x)\,dx[/MATH]
Hence:

[MATH]m=2\int_0^4 \sqrt{4^2-x^2}\rho(x)\,dx[/MATH]
Can you proceed?

Yes, thank you for your help!!