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I obtain different answers depending on the order of integration. I suspect it has to do with the bounds I am setting. The function is f(x,y)=xy^2. I seek to find the volume under a portion of the graph. the region on the xy-plane is a right isosceles triangle one of whose legs connects the points (0,0) and (2,0) on the x-axis, while the other leg connects the points (2,0) and (2,2). 1. When the inner integral is dy, I used 0->x as the bounds..I wrote the upper bounds in terms of x. I obtained antiderivative x^4/3. Then integrated this function in terms of x (dx)->obtained antiderivative of x^5/15 and from 0 to 2, the answer was 32/15 cubic units 2. When the inner integral is dx, I used 0->y as the bounds...I wrote the upper bound in terms of y. I obtained the antiderivative of y^4/2. Then integrated this function in terms of y, dy. --> obtained antiderivative of y^5/10 and from 0 to 2. the answer was 32/10 cubic units.

I think because of the nature of the region on the xy-plane, the bounds are incorrect but I simply cannot reconcile this discrepancy

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Just an arithmetic mistake on the 2nd integral. I get $$ \begin{split} A &= \int_{y=0}^{y=2} \int_{x=y}^{x=2} xy^2dxdy \\ &= \int_0^2 y^2 \left( \int_y^2 xdx \right) dy \\ &= \int_0^2 y^2 \left( \left[\frac{x^2}{2}\right]_y^2=\frac{4-y^2}{2} \right) dy \\ &= \frac{1}{2}\int_0^2 y^2 (4-y^2) dy \\ &= \frac{1}{2}\int_0^2 \left(4y^2 -y^4\right) dy \\ &= \frac{1}{2} \left[\frac{4y^3}{3} - \frac{y^5}{5} \right]_0^2 \\ &= \frac{1}{2} \left(\frac{32}{3} - \frac{32}{5}\right) \\ &= \frac{32}{2} \frac{5-3}{3 \cdot 5} \\ &= \frac{32}{15}. \end{split} $$

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