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Find Volume under surface z(x,y)=x+y and above the triangle defined in x-y plane by 0<=y<=x and 0<=x<=12

asked in Calculus | 524 views
0
864

Volume  is calculate as $\iint \limits_V f(x,y) \,dy \,dx$

so,  $\int^{12}_0\int^x_0 z(x,y) \,dy \,dx$

$=\int^{12}_0\int^x_0 (x+y) \,dy \,dx$

$=\int^{12}_0 (xy+\frac{y^2}{2}) |^x_0 \,dx$

$=\int^{12}_0 (x^2+\frac{x^2}{2}) \,dx$

$=\frac{3}{2}\int^{12}_0 x^2 \,dx$

$=\frac{3}{2}\left ( \frac{x^3}{3} |^{12}_0 \right )$

$=864$

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