Chapter 5: Q22P (page 247)

Above the triangle with vertices (0,2),(1,1) and (2,2) , and under the surface z = xy.

Expert verified

The required solution is $\frac{5}{3}$

Step by step solution

The double integral of f(x,y)over the area Ain the (x,y)plane as the limit of this sum, and we write it as $\int \int_{A} f (x, y) dxdy$

The volume above the triangle with vertices (0,2), (1,1) and (2,2), and under the surface z = xy .

First, integrate over z :

$| = \int_{x = 1}^{2} \int_{y = 2 - x}^{x} [\int_{z = 0}^{xy} dz] dxdy = \int_{1}^{2} dx \int_{2 - x}^{x} dyxy$

Now, integrate over y:

$| = \int_{1}^{2} xdx {(\frac{1}{2} y^{2})}_{2 - x}^{x} = \frac{1}{2} \int_{1}^{2} x (x^{2} - 4 + 4 x - x^{2}) dx$

Now, integrate over x:

$| = 2 {(\frac{x^{3}}{3} - \frac{1}{2} x^{2})}_{1}^{2} = \frac{5}{3}$

Therefore, the value is $\frac{5}{2}$

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