Find the volume using multiple integral in the first octant bounded by the surfaces z = x + y, y = 1 - x^2.

$y_U = 1 - x^2$

$y_L = z - x$

$V = {\displaystyle \int_0^1 \int_0^z} (y_U - y_L) \, dx \, dz$

$V = {\displaystyle \int_0^1 \int_0^z} \left[ \, (1 - x^2) - (z - x) \, \right] \, dx \, dz$

$V = {\displaystyle \int_0^1} \left[ x - \dfrac{x^3}{3} - zx + \dfrac{x^2}{2} \right]_0^z \, dz$

$V = {\displaystyle \int_0^1} \left( z - \frac{1}{3}z^3 - z^2 + \frac{1}{2}z^2 \right) \, dz$

$V = {\displaystyle \int_0^1} \left( z - \frac{1}{3}z^3 - \frac{1}{2}z^2 \right) \, dz$

$V = \left[ \dfrac{z^2}{2} - \dfrac{z^4}{12} - \dfrac{z^3}{6} \right]_0^1$

$V = \frac{1}{2} - \frac{1}{12} - \frac{1}{6}$

$V = \frac{1}{4} ~ \text{unit}^3$

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$y_U = 1 - x^2$

$y_L = z - x$

$V = {\displaystyle \int_0^1 \int_0^z} (y_U - y_L) \, dx \, dz$

$V = {\displaystyle \int_0^1 \int_0^z} \left[ \, (1 - x^2) - (z - x) \, \right] \, dx \, dz$

$V = {\displaystyle \int_0^1} \left[ x - \dfrac{x^3}{3} - zx + \dfrac{x^2}{2} \right]_0^z \, dz$

$V = {\displaystyle \int_0^1} \left( z - \frac{1}{3}z^3 - z^2 + \frac{1}{2}z^2 \right) \, dz$

$V = {\displaystyle \int_0^1} \left( z - \frac{1}{3}z^3 - \frac{1}{2}z^2 \right) \, dz$

$V = \left[ \dfrac{z^2}{2} - \dfrac{z^4}{12} - \dfrac{z^3}{6} \right]_0^1$

$V = \frac{1}{2} - \frac{1}{12} - \frac{1}{6}$

$V = \frac{1}{4} ~ \text{unit}^3$

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