2,314 views
2 votes
2 votes

 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

1 Answer

Best answer
2 votes
2 votes

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

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

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

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

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

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

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

$=864$

selected by

Related questions

0 votes
0 votes
1 answer
1
Jason asked Sep 3, 2018
1,104 views
The area between the parabola $x^2 = 8y$ and the straight line y= 8 is .I am getting $128√2$.
1 votes
1 votes
0 answers
2
Tuhin Dutta asked Jan 25, 2018
596 views
x00.30.60.91.21.51.82.12.4f(x)00.090.360.811.442.253.244.415.76The value of the below integral computed using the continuous at x = 3?$$\int_{0}^{3} f(x) dx$$a) 8.983b) 9...
1 votes
1 votes
1 answer
3
Shubhanshu asked Sep 8, 2017
826 views
Let f(x)=x−(1/2) and A denote the area of region bounded by f(x) and the x-axis, when x varies from -1 to 1.A is nonzero and finite??
1 votes
1 votes
2 answers
4
sh!va asked Mar 10, 2017
976 views
The area bounded by the curves $y^2$ = 9x, x - y + 2 = 0 is given bya) 1b) 1/2C) 3/2d) 5/4