Compute volume below z=x^2+y and above [0,1]x[1,2]

[jonroberts74]

Homework Statement

compute volume below z=x^2+y and above the rectangle R= [0,1]x[1,2]

The Attempt at a Solution

\int_{0}^{1}\int_{1}^{2} x^2+y dydx

\int_{0}^{1} [x^2y+\frac{y^2}{2}]\Bigg|_{1}^{2} dx

\int_{0}^{1} x^2+ \frac{3}{2}dx

\frac{x^3}{3} + \frac{3x}{2} \Bigg|_{0}^{1} = \frac{11}{6}

checking to make sure I understood the question correctly

 

[ehild]

Homework Statement

compute volume below z=x^2+y and above the rectangle R= [0,1]x[1,2]

The Attempt at a Solution

\int_{0}^{1}\int_{1}^{2} x^2+y dydx

\int_{0}^{1} [x^2y+\frac{y^2}{2}]\Bigg|_{1}^{2} dx

\int_{0}^{1} x^2+ \frac{3}{2}dx

\frac{x^3}{3} + \frac{3x}{2} \Bigg|_{0}^{1} = \frac{11}{6}

checking to make sure I understood the question correctly

It is all rigth, but do not forget the parentheses next time.

\int_{0}^{1}\int_{1}^{2}{( x^2+y )dydx}

\int_{0}^{1}{ (x^2+ \frac{3}{2})dx}

ehild

 

[HallsofIvy]

Since x and y happen to be "separated" in this integral, you could also do it as
\int_0^1\int_1^2 (x^2+ y) dy dx= \int_0^1\int_1^2 x^2 dydx+ \int_0^1\int_1^2 y dy dx
= \left(\int_0^1 x^2 dx\right)\left(\int_1^2 dy\right)+ \left(\int_0^1 dx\right)\left(\int_1^2 y dy\right)= \left[\frac{1}{3}x^2\right]_0^1\left[y\right]_1^2+ \left[x\right]_0^1\left[\frac{1}{2}y^2\right]_1^2= \left[\frac{1}{3}\right]\left[1\right]+ \left[1\right]\left[\frac{3}{2}\right]= \frac{11}{6}