Q: Determine the mass and center of mass of the lamina for region D if D is bounded by \(\displaystyle y=x^2\) and \(\displaystyle x=y^2\); \(\displaystyle p(x,y)=5\sqrt{x}\)

Could someone assist me with setting up the limits on the integral for finding the mass?

Well,

\(\displaystyle C.O.M = ( \bar x , \bar y ) \)

Where,

\(\displaystyle \bar x = \frac{ \iint_D x P(x,y) dA }{ \iint_D P(x,y) dA} \)

\(\displaystyle = \frac{ 5 \int_0^1 x \sqrt{x} dx \int_{x^2}^{ \sqrt{x} }dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy } \)

\(\displaystyle = \frac{ 5 \int_0^1 x \sqrt{x} ( \sqrt{x} - x^2 ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx } \)

\(\displaystyle = \frac{ 5 \int_0^1 (x^2 - x^{ \frac{7}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx } \)

And,

\(\displaystyle \bar y = \frac{ \iint_D y P(x,y) dA }{ \iint_D P(x,y) dA} \)

\(\displaystyle = \frac{ 5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } y dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy } \)

\(\displaystyle = \frac{ 5 \int_0^1 \sqrt{x} ( \frac{ x^2 }{2} - \frac{x^4 }{2} ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx } \)

\(\displaystyle = \frac{ \frac{5}{2} \int_0^1 ( x^{ \frac{5}{2} } - x^{ \frac{9}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx } \)

If you don't know double integrals yet that's fine, just compute them as the single integrals that I've reduced them to. This will serve as your C.O.M.

Your mass, as Skeeter pointed out is the denominator in the above calculations. So,

\(\displaystyle Mass = 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx \)