Question

Let Ω be the region above y+z= 2, below z= 4, and between x= 0 and x= 4−y^2 a triple integral for the volume of the region. Set up an integral to represent the volume of Ω in all six orders and then fully solve ONE of them.

Answer 1

There are 6 possible ordersof integration

$$dzdxdy,dzdydx,dydzdx,dydxdz,dxdydz,dxdzdy$$

$$V=\iiint_\Omega dV$$

$$-2\leq y\leq 2, \ 0\leq x\leq 4-y^2,\ 2-y\leq z\leq 4$$

$$V=\displaystyle\int_{-2}^2\displaystyle\int_{0}^{4-y^2}\displaystyle\int_{2-y}^4dzdxdy=$$

$$=\displaystyle\int_{-2}^2\displaystyle\int_{0}^{4-y^2}[z]\begin{matrix} 4 \\ 2-y \end{matrix}dxdy=\displaystyle\int_{-2}^2\displaystyle\int_{0}^{4-y^2}(2+y)dxdy=$$

$$=\displaystyle\int_{-2}^2[2x+xy]\begin{matrix} 4-y^2 \\ 0 \end{matrix}dy=\displaystyle\int_{-2}^2\big(2(4-y^2)+y(4-y^2)-0\big)dy=$$

$$=\bigg[ 8y-{2y^3 \over 3}+2y^2-{y^4 \over 4}\bigg]\begin{matrix} 2 \\ -2 \end{matrix}=$$

$$=16-{16 \over 3}+8-4+16-{16 \over 3}-8+4={64 \over 3}(units^3)$$

$$0\leq x\leq 4, \ -\sqrt{4-x}\leq y\leq \sqrt{4-x},\ 2-y\leq z\leq 4$$ $$V=\displaystyle\int_{0}^4\displaystyle\int_{-\sqrt{4-x}}^{\sqrt{4-x}}\displaystyle\int_{2-y}^4dzdydx$$

$$-2\leq y\leq 2, \ 2-y\leq z\leq 4,\ 0\leq x\leq 4-y^2$$ $$V=\displaystyle\int_{-2}^2\displaystyle\int_{2-y}^4\displaystyle\int_{0}^{4-y^2}dxdzdy$$

$$0\leq z\leq 4, \ -2\leq y\leq 2-z,\ 0\leq x\leq 4-y^2$$ $$V=\displaystyle\int_{0}^4\displaystyle\int_{-2}^{2-z}\displaystyle\int_{0}^{4-y^2}dxdydz$$

$$0\leq z\leq 4, \ 0\leq x\leq 4-(2-z)^2,\ -2\leq y\leq 2-z$$ $$V=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4-(2-z)^2}\displaystyle\int_{-2}^{2-z}dydxdz$$

$$0\leq x\leq 4, \ 2-\sqrt{4-x}\leq z\leq 2+\sqrt{4-x},\ -2\leq y\leq 2-z$$ $$V=\displaystyle\int_{0}^4\displaystyle\int_{2-\sqrt{4-x}}^{2+\sqrt{4-x}}\displaystyle\int_{-2}^{2-z}dydzdx$$