Answer to Question #254217 in Calculus for Susan

Question

Find the volume of the region below the surface f(x,y) = 3y^2 + x and above the region of the xy-plane enclosed by the lines y = 2, x=0, y = 0, and y = x - 1.

Answer 1

Let us find the volume of the region below the surface $f(x,y) = 3y^2 + x$ and above the region of the xy-plane enclosed by the lines $y = 2, x=0, y = 0,$ and $y = x - 1.$

$V=\int_{-1}^2 dy\int_0^{y+1} f(x,y)dx =\int_{-1}^2 dy\int_0^{y+1} (3y^2 + x)dx =\int_{-1}^2 dy (3y^2x + \frac{x^2}2)|_0^{y+1}\\ =\int_{-1}^2 (3y^2(y+1)+\frac{(y+1)^2}2)dy =\int_{-1}^2 (3y^3+3y^2+\frac{y^2+2y+1}2)dy =\int_{-1}^2 (3y^3+\frac{7}2y^2+y+\frac{1}2)dy\\ =(\frac{3}4y^4+\frac{7}6y^3+\frac{y^2}2+\frac{1}2y)|_{-1}^2 =12+\frac{28}3+2+1-(\frac{3}4-\frac{7}6+\frac{1}2-\frac{1}2) =15+\frac{28}3-\frac{3}4+\frac{7}6 =15+\frac{112-9+14}{12} =15+\frac{117}{12}=15+\frac{39}4=24.75$(cubis units)