Answer
$m$ = $\frac{8k}{15}$
$(0,\frac{4}{7})$
Work Step by Step
$m$ = $\int_{-1}^1\int_0^{1-x^2}kydydx$
$m$ = $\frac{k}{2}\int_{-1}^1[y^2]|_0^{1-x^2}dx$
$m$ = $\frac{k}{2}\int_{-1}^1[1-2x^2+x^4]dx$
$m$ = $\frac{k}{2}[x-\frac{2x^3}{3}+\frac{x^5}{5}]_{-1}^1$
$m$ = $\frac{8k}{15}$
$x̄$ = $\frac{1}{m}\int_{-1}^1\int_0^{1-x^2}kxydydx$
$x̄$ = $\frac{15}{8}\int_{-1}^1\int_0^{1-x^2}xydydx$
$x̄$ = $\frac{15}{8}\int_{-1}^1[\frac{xy^2}{2}]|_0^{1-x^2}dx$
$x̄$ = $\frac{15}{16}\int_{-1}^1(x-2x^3+x^5)dx$
$x̄$ = $\frac{15}{16}[\frac{x^2}{2}-\frac{x^4}{2}+\frac{x^6}{6}]_{-1}^1$
$x̄$ = $0$
$ȳ$ = $\frac{1}{m}\int_{-1}^1\int_0^{1-x^2}ky^2dydx$
$ȳ$ = $\frac{15}{8}\int_{-1}^1\int_0^{1-x^2}y^2dydx$
$ȳ$ = $\frac{15}{8}\int_{-1}^1[\frac{y^3}{3}]_0^{1-x^2}dx$
$ȳ$ = $\frac{5}{8}\int_{-1}^1(1-3x^2+3x^4-x^6)dx$
$ȳ$ = $\frac{5}{8}[x-x^3+\frac{3x^5}{5}-\frac{x^7}{7}]_{-1}^1$
$ȳ$ = $\frac{4}{7}$
