Question

2. a) Calculate the volume under the plane z=x+2y and over the region R={(x,y)|1≤x≤2 and 3≤y≤5}.
b) Calculate the volume under the surface z=xy^2+y^3 and the region R={(x,y)|0≤x≤2 and 1≤y≤3},

Answer 1

If z=f(x,y) - surface, then volume under this surface is calculated as $\int_R f(x,y)dxdy$ where R is a region of xOy plane.

2.a) $\int_1^2dx\int_3^5(x+2y)dy=\int_1^2(xy+y^2)|_3^5 dx=$

$=\int_1^2(5x+25-3x-9) dx=\int_1^2(2x+16) dx=$

$=(x^2+16x)|_1^2=4+32-1-16=19$ .

Answer: 19

2.b) $\int_0^2dx\int_1^3(xy^2+y^3)dy=\int_0^2(\frac{xy^3}{3}+\frac{y^4}{4})|_1^3dx=$

$=\int_0^2(9x+\frac{81}{4}-\frac{x}{3}-\frac{1}{4})dx=\int_0^2(\frac{26x}{3}+20)dx=$

$=(\frac{13x^2}{3}+20x)|_0^2=\frac{52}{3}+40=\frac{172}{3}=57\frac{1}{3}$ .

Answer: $57\frac{1}{3}$