A lamina occupies the part of the disk $x^2+y^2\leq 25$ in the first quadrant.

Here is the question: A lamina occupies the part of the disk $$x^2+y^2\leq 25$$ in the first quadrant. Its density is $$\rho=4$$.

1. What is the total mass?
2. What is the moment about the x-axis?
3. What is the moment about the y-axis?

Thanks a ton.

• Welcome to MSE! It is best practice to show what you have tried when asking for help. Where are you stuck? For total mass, you just need to multiply the density by the area of the region. There are also formulas for the $x,y$ moments. – D.B. Mar 29 at 17:49

\begin{align*} M = \iint_{R}\rho(x,y)\mathrm{d}y\mathrm{d}x = 4\int_{0}^{5}\int_{0}^{\sqrt{25-x^{2}}}1\mathrm{d}y\mathrm{d}x \end{align*}
For the second question, consider \begin{align*} I_{x} = \iint_{R}r^{2}\mathrm{d}m = \iint_{R}y^{2}\rho(x,y)\mathrm{d}y\mathrm{d}x = 4\int_{0}^{5}\int_{0}^{\sqrt{25-x^{2}}}y^{2}\mathrm{d}y\mathrm{d}x \end{align*}
Finally, for the last question, we have \begin{align*} I_{y} = \iint_{R}r^{2}\mathrm{d}m = \iint_{R}x^{2}\rho(x,y)\mathrm{d}y\mathrm{d}x = 4\int_{0}^{5}\int_{0}^{\sqrt{25-x^{2}}}x^{2}\mathrm{d}y\mathrm{d}x \end{align*}