ok so the question asks for the volume of the function
$z = 1-xy$ on the first octant bounded by
$x = y$ and $x = y^2 $
i know that on the interval $x = y$ is greater than $x = y^2$ so the limits look like $$\int_0^1\int_{y^2}^y$$ (the intersection on first octant) NOTE: my integration using dA = dx dy however, when i evaluate this iterated integral i am returned with $V=-0.5417$ ?
ive checked the graph of this and it looks like the volume should be positive with the given boundaries. I know that double integrals are not always equal to the iterated but my prof said for her questions it should be. What am i doing wrong? is the outer bound supposed to be from 1 to 0 instead to yield a positive volume, but how come thats the case?
here's my work but idk how to format integral symbol on here V = integral form 0 to 1 of (-y - ((y^3)/2) + ((y^5)/2) )dy which gives V = -0.5417
$$ \overbrace{\left(\int_0^1 dx \int_{x^2}^x dy\ 1\right)}^A - \overbrace{\left(\int_0^1 dx \int_{x^2}^x dy\ xy \right)}^B $$
$$ A = \int_0^1 (x -x^2)\ dx = \left[\frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac12 - \frac13 $$ $$ B = \int_0^1 x \frac12 (x^2 - x^4)\ dx = \frac12 \left[\frac{x^4}{4} -\frac{x^6}{6} \right]_0^1 = \frac12(\frac14 - \frac16) $$ Combining all gives you $\frac18$