1) Estimate the volume of the solid that lies below the surface z = x + 2y^2 and

Question

1) Estimate the volume of the solid that lies below the surface z = x + 2y^2 and above the rectangle R = [0,3] * [0,4] with m = 3 and n = 2. Choose the sample points to be the lower right corners. 2) Calculate the double integral by first reversing the order of integration. intergral 0 to1 , integral 0 to cos^-1 y . sinx sqroot(sin^(2)x + 1)dxdy 3) Find the volume of the solid bounded by the cylinder x^2 + y^2 = 9, the plane z = 0, and the surface z = xy in the first octant by using rectangular coordinates.

Explanation / Answer

Lower right means lower value for y and right value for x.

Thus, [0,3] has m=3, and (3-0)/3 = 1, so we find the values on the right side at 3, 2, and 1

[0,4] has n = 2, and (4-0)/2 = 2, so bottom values are 0 and 2.

Finally, the area of each rectangle is 1x2 = 2

Thus, the estimated volume is 2(f(1,0) + f(1,2) + f(2,0) + f(2,2) + f(3,0) + f(3,2)) =

2(1+2*0^2 + 1 + 2*2^2+2+2*0^2 + 2 + 2*2^2+3+2*0^2 + 3 + 2*2^2) =

2(2*1+2*2+2*3 + 3 *2*0^2 +3*2*2^2) =

2(2+4+6 + 0 + 24) =

2(36) =

72

2) Originally x ranges from 0 to arccos y, while y ranges from 0 to 1. Reversing the order of integration, note that x = arccos(y) when y = cos(x) Thus, let x range from 0 to pi/2, and y range from 0 to cos x

Thus, using I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b, we have

I[0, pi/2] I[0, cos x] sinx sqroot(sin^(2)x + 1) dy dx =

I[0, pi/2] sinx sqroot(sin^(2)x + 1) y E[0, cos x] dx =

I[0, pi/2] sinx sqroot(sin^(2)x + 1)(cosx-0) dx =

I[0, pi/2] sinx cos x sqroot(sin^(2)x + 1) dx =

1/3 (sin^(2)x + 1)^(3/2) E[0, pi/2] =

1/3(sin^(2)pi/2+1)^(3/2) - 1/3(sin^(2)0+1)^(3/2) =

1/3*2^(3/2) - 1/3*1^(3/2) =

2/3 sqrt(2) - 1/3

3)

In rectangular coordinates, let z range from 0 to xy.

Let x range from 0 to sqrt(9-y^2), and y range from 0 to 3,

as x^2+y^2 <= 9 and we are in the first quadrant.

Then, I[0,3] I[0, sqrt(9-y^2)] I[0, xy] 1 dz dxdy =

I[0,3] I[0, sqrt(9-y^2)] z E[0, xy] dxdy =

I[0,3] I[0, sqrt(9-y^2)] xy - 0 dxdy =

I[0,3] I[0, sqrt(9-y^2)] xy dxdy =

I[0,3] 1/2x^2y E[0, sqrt(9-y^2)] dy =

I[0,3] 1/2(sqrt(9-y^2)^2-0^2)y dy =

I[0,3] 1/2(9-y^2)y dy =

I[0,3] 1/2(9y-y^3) dy =

9/4y^2 - y^4/8 E[0,3] =

9/4*9 - 81/8 =

81/8

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