Query 125

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course Mth 277

4/18

Question: `q001. Compute the iterated triple integral Int( Int( Int(x^2*y sin(xyz) dz, 0,1) dy,0,1) dx,0, pi)

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Your solution:

Int( Int( Int(x^2*y sin(xyz) dz, 0,1) dy,0,1) dx,0, pi)

Int(x^2*y sin(xyz) dz) = (x^2*y)Int(sin(xyz) dz) = (x^2*y)*(-cos(xyz)/xy) = -xcos(xyz), evaluating

(-xcos(xy*1)) - (-xcos(xy*0)) = x - xcos(xy)

Int( Int( x - xcos(xy) dy,0,1) dx,0, pi)

Int( x - xcos(xy) dy) = Int( x dy) - Int( xcos(xy) dy)

Int( x dy) = xy

Int(xcos(xy) dy) = x*Int(cos(xy)) dy) = x(sin(xy)/x) = sin(xy)

Int( x - xcos(xy) dy) = xy - sin(xy), evaluating

[ x*1 - sin(x*1)] - [x*0 - sin(x*0)] = (x - sin(x)) - (0) = x-sin(x)

Int(x-sin(x) dx,0,pi)

Int(x-sin(x) dx) = Int( x dx) - Int( sin(x) dx)

Int(x dx) = x^2/2

Int(sin(x) dx) = -cos(x)

Int(x-sin(x) dx) = x^2/2 + cos(x), evaluated

[`pi^2/2 + cos(`pi)] - [0^2/2 + cos(0)] = `pi^2/2 - 2 = approx. 2.935

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Given Solution:

The inner integrand is x^2 y sin(x y z) with respect to z. Antiderivative is -x^2 y * (cos(x y z) / x y) = x cos(x y z). Between z = 0 and z = 1 this changes from -x cos(0) to -x cos(x y), a change of x - x cos(x y).

The middle integrand is therefore x - x cos(x y), integrated with respect to y. Antiderivative is - x sin(x y) / x + x y / 2 = -sin(x y) + xy. Between y = 0 and y = 1 this changes from - sin(0) + 0 to -sin(x) + x, a change of -sin(x) + x.

The outer integrand is therefore - sin(x) + x = cos(x) + x^2 / 2. The change in the value of the antiderivative between x = 0 and x = pi is thus cos(pi) + pi^2 / 2 - cos(1) = -2 + pi^2 / 2.

In standard form the integral is written

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Self-critique (if necessary):

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Question: `q002. Evaluate the triple integral of xz + 2yx over D with respect to V where D is the box 2<= x <= 4, 1 <= y <= 3, -1 <= z <= 1

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Your solution:

Int( Int( Int(xz + 2yx dz, -1,1) dy,1,3) dx,2,4)

Int(xz + 2yx dz ) = (x)Int( z dz) + (2xy)*Int(1 dz)

(x)Int( z dz) = x(z^2/2)

(2xy)*Int(1 dz) = (2xy)*(z) = 2xyz

Int( x - xcos(xy) dy) = x(z^2/2) + 2xyz, evaluating

[ x(1^2/2) + 2xy*1] - [x(-1^2/2) + 2xy*(-1)] = 4xy

Int( Int( 4xy dy,1,3) dx,2, 4)

Int( 4xy dy) =4x* Int( y dy) = 4x*(y^2/2), evaluating

[ 4x*(3^2/2),] - [4x*(1^2/2),] = 18x - 4x = 14x

Int(14x dx,2,4)

Int(14x dx) = 14*Int( x dx) = 14*(x^2/2), evaluated

[14*(4^2/2) ] - [14*(2^2/2)] = 14*(8 - 2) = 14*6 = 84

Int( Int( Int(xz + 2yx dz, -1,1) dy,1,3) dx,2,4) = 84

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Given Solution:

The integral is

int(int(int((xz+2yx) dz, -1, 1) dy, 1, 3) dx, 2, 4)

The inner integrand is x z + 2 y x, integrated with respect to z, which gives us antiderivative x z^2 / 2 + 2 y x z. Between z = -1 and 1 this antiderivative changes from x / 2 - 2 y x to x / 2 + 2 y x = 4 y x.

The middle integrand is therefore 4 y x with respect to y, which gives us antiderivative 2 y^2 x. Between y = 1 and 3 this antiderivative changes from 2 x to 18 x, a change of 16 x.

The outer integrand 16 x is easily evaluated between x = 2 and x = 4. Antiderivative is 8 x^2, which changes from 32 to 128, a change of 96.

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Self-critique (if necessary):

I did not divide 4x by 2 to get 2x, so when evaluated instead of 16x I was working with 14x.

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Question: `q003. Use a triple integral to find the volume of the solid bounded by y = 4 - x^2, z = 0, and z = y.

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Your solution:

For y = 4-x^2, we have a parabola with orgin at (0,4) in the xy plane, so y has domain of 4 to -infinity but because we are looking a in the boundary of z=0 and z = y, z then has a max of 4(max value of y) and we are only looking at the parabola from y=4 to x axis(y=0) so our integral is

Int(Int(Int( dz,0,4) dy,0,4-x^2) dx,-2,2)

Int( dz) = z,evaluated

4 - 0 = 4

Int( 4 dy) = 4y,evaluated

4*(4-x^2) = 16x^2

Int(16 - 4x^2 dx) = 16x - (4/3)x^3,evaluated

[32 - (4/3)*8] - [32 -(4/3)8] = 64

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Given Solution:

y = 4 - x^2 is a parabolic cylinder extending above and below the parabola y = 4 - x^2 in the x y plane. This parabola has vertex at (0, 4) and opens in the negative y direction, passing through the x axis at (2, 0) and (-2, 0).

The plane z = 0 is just the x y plane, and the plan z = y 'slices' 3-dimensional space at an angle of 45 degrees to the x y plane, 'cutting through' at the x axis.

The region of the cylinder above the x y plane and below the z = y plane is finite. The region above the z = y plane and below the x y plane is infinite in extent.

We integrate to find the volume of the finite region.

In the x y plane the region beneath the cylinder can be described as -2 <= x < = 2, 0 <= y <= 4 - x^2, 0 <= z <= y.

Our integral is therefore

int(int(int(dz, 0, y) dy, 0, 4 - x^2), dx, -2, 2).

Our inner integrand is just 1 with respect to z, with antiderivative z. Between z = 0 and z = y our antiderivative changes by y.

The integrand of our middle integral is therefore y, integrated with respect to y, giving us antiderivative y^2 / 2. Between y = 0 and y = 4 - x^2 this antiderivative changes by (4 - x^2)^2 / 2 = (16 - 8 x^2 + x^4) / 2 = 8 - 4 x^2 + x^4/2.

The integrand of our outer integral is therefore 8 - 4 x^2 + x^4/2, integrated with respect to x. Our antiderivative is 8 x - 4/3 x^3 + x^5 / 10. Between -2 and 2 this antiderivative changes by 2 * (16 - 4/3 * 8 + 32/10) = 256/15.

It might be instructive to contrast this with the double integral which yields the same result.

For each area increment `dA with sample point (x_hat, y_hat), the corresponding region of our solid extends from z = 0 to z = y_hat. So the volume of the region is y_hat * `dy * `dx.

The resulting Riemann sum approaches the integral int(int(y dy, 0, 4-x^2) dx, -2, 2) = 256/15.

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Self-critique (if necessary):

I put z boundary from 0 to 4 but I guess it should have been from 0 to y. I believe that was my only mistake.

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Question: `q004. Change the order of integration of the triple integral Int(Int(Int(f(x,y,z) dz, 0, 1-2x) dy, 0, 1-4x^2) dx, 0, 1/2) to dy dx dz.

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Your solution:

Everything is given in terms of x so we have

z = 1-2x

y = 1-4x^2

Solving for x, respectively,

x = z/2 -1

x = `sqrt((y-1)/4)

Looking at graph of y = 1-4x^2, where x is from 0 to (1/2) as in original boundary the integral is aimed to find the area within the first Quad and we can see that where x runs from 0 to (1/2) that y runs from 0 to 1 during the interval defined by y = 1-4x^2 and for 0<=x<=(1/2) so now we put together our integral

Int(Int(Int( f(x,y,z) dy,0,1) dx,0,`sqrt((y-1)/4)) dz,0, `sqrt((y-1)/4))

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Given Solution:

The region can be described by 0 <= x <= 1/2, 0 <= y <= 1 - 4 x^2, 0 <= z <= 1 - 2 x.

0 <= x <= 1/2, 0 <= y <= 1 - 4 x^2 describes the region between the x axis and the parabola y = 1 - 4 x^2. The vertex is (0, 1) and the x intercepts are (+- 1/2, 0).

This region can also be described as follows:

For any y between 0 and 1, the horizontal line through (0, y) meets the boundary curve at coordinate x, such that y = 1 - 4 x^2. The value of x is therefore x = +- sqrt( (1 - y) / 4) = +- 1/2 sqrt( 1 - y ). The horizontal line is inside the region at the point (0, y) and therefore remains inside the region between these two values of x.

The region in the x y plane can therefore be described as 0 <= y <= 1/2, -1/2 sqrt( 1 - y ) <= x <= 1/2 sqrt( 1 - y ).

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Self-critique (if necessary):

??Why is the defined region in the x y plane described as 0 <= y <= 1/2, -1/2 sqrt( 1 - y ) <= x <= 1/2 sqrt( 1 - y ) and not 0<=y<=1?????

@& That was probably a typo. Should be 0 <= y <= 1.*@

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Question: `q005. Use triple integration to find the volume of the following solids.

• A sphere of radius r.

• An ellipsoid with equation x^2/a^2 + y^2/b^2 + z^2/c^2 = 1 (a,b,c > 0).

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Your solution:

A sphere of radius r.

for sphere of radius r we have in xy-plane we have x^2+y^2 +z^2 = r^2, where -r<=x<=r and -`sqrt(r^2 - x^2)<=y<=`sqrt(r^2-x^2) so we get

Int(Int(Int( f(x,y,z) dz, -`sqrt(r^2-x^2-y^2), `sqrt(r^2-x^2-y^2)) dy, -`sqrt(r^2-x^2), -`sqrt(r^2-x^2)) dx,-r,r)

????I’m confused about how to set these up????

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Given Solution:

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Self-critique (if necessary):

@& For example consider a sphere of radius 1 centered at the origin.

It lies above the circle of radius 1 in the xy plane.

That circle is defined by

x^2 + y^2 = 1.

The region inside the circle is bounded by y = sqrt(1 - x^2) and y = - sqrt(1 - x^2), and can therefore be described by

-1<= x <= 1

-sqrt(1 - x^2) <= y <= sqrt(1 - x^2).

Now in the x y z plane, any point (x, y) within the above circle is bounded above and below by the sphere.

The sphere is characterized by

x^2 + y^2 + z^2 = 1

so the z coordinates above and below (x, y) are

z = sqrt( 1 - x^2 - y^2)

and

z = - sqrt( 1 - x^2 - y^2).

Thus the description of the region is

-1<= x <= 1

-sqrt(1 - x^2) <= y <= sqrt(1 - x^2)

=sqrt( 1 - x^2 - y^2) <= z <= sqrt( 1 - x^2 - y^2)

See if this helps. If not, ask more questions.*@

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&#This looks good. See my notes. Let me know if you have any questions. &#