Find the volume of the given solid. bounded by the planes z=x,y=x,x+y=9 and z=0 ◻

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Sketch Region and Bounds: Sketch the region and identify the bounds. The solid is bounded by the planes z=x, y=x, x+y=9, and z=0. The intersection of the planes y=x and x+y=9 is a line in the xy-plane, which we can find by solving the system of equations. y = x x + y = 9 Substitute y=x into the second equation: x + x = 9 2x = 9 x = 9/2 y = 9/2 So, the line of intersection is at the point (9/2, 9/2) in the xy-plane.Set Up Double Integral: Set up the double integral for the volume. The volume V of the solid can be found by integrating over the region in the xy-plane. The bounds for y are from y=x to y=9-x, and the bounds for x are from x=0 to x=9/2. V = ∫∫(z) dy dx, where z ranges from 0 to x.Compute Double Integral: Compute the double integral. V = ∫ from x=0 to x=9/2 (∫ from y=x to y=9-x (x) dy) dx First, integrate with respect to y: V = ∫ from x=0 to x=9/2 (x*(9-x) - x*x) dx V = ∫ from x=0 to x=9/2 (9x - x^2 - x^2) dx V = ∫ from x=0 to x=9/2 (9x - 2x^2) dxContinue Integration: Continue the integration with respect to x. V = [ (9/2)x^2 - (2/3)x^3 ] from x=0 to x=9/2 Now, plug in the upper and lower bounds of x: V = (9/2)*(9/2)^2 - (2/3)*(9/2)^3 - ( (9/2)*0^2 - (2/3)*0^3 ) V = (9/2)*(81/4) - (2/3)*(729/8) V = (729/8) - (2/3)*(729/8) V = (729/8) * (1 - 2/3) V = (729/8) * (1/3) V = 729/24 V = 30.375

Find the volume of the given solid. $\newline$ bounded by the planes $z=x, y=x, x+y=9$ and $z=0$ $\newline$ $\square$

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Find the volume of the given solid.\newlinebounded by the planes z=x,y=x,x+y=9 z=x, y=x, x+y=9 z=x,y=x,x+y=9 and z=0 z=0 z=0\newline□ \square □

Find the volume of the given solid. $\newline$ bounded by the planes $z=x, y=x, x+y=9$ and $z=0$ $\newline$ $\square$