Need Help?

  • Notes
  • Comments & Questions

Evaluate the iterated integral.
$$\int_{0}^{2} \int_{0}^{z^{2}} \int_{0}^{y-z}(2 x-y) d x d y d z$$

$$\int_{0}^{2} \int_{0}^{z^{2}} \int_{0}^{y-z}(2 x-y) d x d y d z$$

$$= \int_{0}^{2} \int_{0}^{z^{2}}\left[\int_{0}^{y-z} x^{2}-x y\right] d y d z$$

$$=\int_{0}^{2} \int_{0}^{z^{2}}\left[(y-z)^{2}-(y-z) y\right] d y d z$$

$$\int_{0}^{2} \int_{0}^{z^{2}}\left(z^{2}-y z\right) d y d z$$

$$\int_{0}^{2}\left[|_{0}^{z^{2}} z^{2} y-\frac{y^{2}}{2} z\right] d z$$

$$\int_{0}^{2} \left[z^{4}-\frac{z^{5}}{2}\right] d z$$

$$=|_{0}^{2}\left[\frac{z^{5}}{5}-\frac{z^{6}}{6 \times 2}\right]$$

$$=\frac{(2)^{5}}{5}-\frac{(2)^{6}}{12}=\frac{16}{15} $$

No comments yet

Join the conversation

Join Notatee Today!