## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{8}{3}$
The average height is equal to: $\dfrac{1}{4} \iint_{R} f dA=\dfrac{1}{4} \times \int^{2}_{0} \int^{2}_{0}(x^2+y^2) \space dy \space dx$ or, $=\dfrac{1}{4} \times \int^{2}_{0}[x^2y+\dfrac{y^3}{3}]^2_0dx$ or, $=\dfrac{1}{4} \times \int^{2}_{0}(2x^2+\dfrac{8}{3}) \space dx$ or, $=\dfrac{1}{2}[\dfrac{1}{3} x^3+\dfrac{4}{3}x]^2_0$ or, $=\dfrac{8}{3}$