calc 3: triple integrals

[rich sentinel]

so the x-part goes from 0 to 3, the z part goes from 0 to $\sqrt{9-y^2}$ and the y part goes from 0 to $\frac{y}{3}$ (the reason y it's dx dz dy instead of dx dy dz is because it's essentially a circle being projected onto the yz plane) and I trust u to go through this for urself and get an answer in the end

[fast gazelle]

madlad-mathemagician

[rich sentinel]

$\int_{0}^{3} \int_{0}^{\sqrt{9-y^2}} \int_{0}^{\frac{y}{3}} dx dz dy$

[fast gazelle]

madlad-mathemagician