## Thomas' Calculus 13th Edition

a) $V=\int_{-3}^{2} \int_{x}^{6-x^2} x^2 \ dy \ dx$ b) $V=\int_{-3}^{2} \int_{x}^{6-x^2} \int_{0}^{x^2} \ dz \ dy \ dx$ c) $\dfrac{125}{4}$
We are given the conditions $x^2+y=6 \implies y=6-x^2$ and $x=y$ a) The double integration for volume can be calculated as: $V=\int_{-3}^{2} \int_{x}^{6-x^2} x^2 \ dy \ dx$ b) The triple integration for volume can be calculated as: $V=\int_{-3}^{2} \int_{x}^{6-x^2} \int_{0}^{x^2} \ dz \ dy \ dx$ c) The volume for double integration can be calculated as: $V=\int_{-3}^{2} \int_{x}^{6-x^2} x^2 \ dy \ dx....(1)$ First we solve $I=\int_{x}^{6-x^2} x^2 \ dy =x^2(6-x^2) -x^3 =[x^2y]_{x}^{6-x^2}=x^2(6-x^2)-x^3$ Now, equation (1) becomes: $V=\int_{-3}^{2} x^2(6-x^2)-x^3 \ dx \\=\int_{-3}^{2} x^2(6-x^2) \ dx -\int_{-3}^{2} x^3 \ dx \\=\int_{-3}^{2} 6x^2-x^4 \ dx -[\dfrac{x^4}{4}]_{-3}^{2}\\=[\dfrac{6x^3}{3}-\dfrac{x^5}{5}]_{-3}^2 -\dfrac{1}{4}[2^4-(-3^4)] \\=15-\dfrac{65}{4} \\=\dfrac{125}{4}$