Query124Mth

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course Mth 277

Question: `q001. Find the surface area of the portion of the surface z = y^2 that lies over the triangular region in the plane with vertices (0,0,0), (1,0,0) and (0,1,0).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Int(Int( `sqrt(f_x + f_y + 1) dy) dx)

f_y = 2y

f_x = 0

Int(Int( `sqrt(2y + 1) dy,0,x-1) dx,0,1)

Int( `sqrt(2y + 1) dy) = (2/3)*(2y + 1)^(3/2)

[(2/3)* (2(x-1) + 1)^(3/2)] - [(2/3)*(2*0 + 1)^(3/2)]

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

Int((2/3)* (2x - 1)^(3/2)- (2/3) dx) = Int((2/3)* (2x - 1)^(3/2) dx) - Int( (2/3) dx)

Int((2/3)* (2x - 1)^(3/2) dx) = (2/3)*Int((2x - 1)^(3/2) dx) = (2/3)*2(2x-1)^(5/2)/5

= 4(2x-1)^(5/2)/15

Int((2/3)dx)=(2/3)Int(dx)=(2/3)x

Int((2/3)*(2x-1)^(3/2)-(2/3)dx)=4(2x-1)^(5/2)/15-(2/3)x

[4(2*1-1)^(5/2)/15-(2/3)*1]-[4(2*0-1)^(5/2)/15-(2/3)*0]=[(4/15)-(2/3)]-(-4/15)=-2/15

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Given Solution:

The triangular region is in the x y plane and is bounded by x axis, the y axis and the line y = 1 - x.

It can therefore be described by 0 <= x <= 1, 0 <= y <= 1 - x.

Quick solution:

We integrate sqrt(1 + f_x^2 + f_y^2) over the region.

For f(x, y) = 2 y, we have

sqrt(1 + f_x^2 + f_y^2) = sqrt( 1 + 4 y^2)

so our integral is

INT(INT(sqrt(1 + 4 y^2) dy, 1, 1 - x), dx, 0, 1),

which is approximately .63.

This makes sense, because the area of the triangular region is .5, and the parabolic region makes a relatively modest angle with vertical above most of the triangle.

Above the point (x, y), a vector normal to the surface z = y^2 is the cross product of i + f_x k and j + f_y k, where f(x, y) = y^2. This gives us the normal vector i X (j + 2 y k) = k - 2 y i.

Alternatively the surface z = f(x, y) is a level surface of the function F(x,y,z) = z - f(x, y). Since the gradient of this function is normal to the level surface, the vector del F(x, y, z) is normal to the surface.

In this case the surface is a level curve of F(x, y, z) = z - f(x, y) = z - y^2, so F_x = 0, F_y = -2 y and F_z = 1. The gradient is therefore 0 i - 2 y j + k., the same as that found previously.

In general the gradient of z - f(x, y) is -f_x i - f_y j + k, and the cosine of its angle with the k vector is 1 / sqrt( 1 + f_x^2 + f_y^2 ).

The unit normal to the xy plane is k, so the cosine of the angle between the normal to the surface and the normal to the xy plane is

cos(theta) = ( k - 2 y i ) dot k / | k - 2 y i | = 1 / sqrt( 1 + (2y)^2 ) = 1 / sqrt( 1 + 4 y^2 ).

The portion of the our surface which lies above an area increment `dA with dimensions `dx by `dy, located at (x_hat, y_hat) has an area that is equal to `dA, if the surface is parallel to the x y plane, and an area greater than `dA, if the surface is not parallel to the x y plane. The ratio of the areas is 1 / cos(theta) = sec(theta), where theta is the angle between the normal to the surface and the normal to the plane. The cosine of that angle was found above to be 1 / sqrt( 1 + 4 y^2 ). The reciprocal is sec(theta) = sqrt( 1 + 4 y^2 ). For the sample point (x_hat, y_hat) we have sec(theta) = sqrt( 1 + 4 y_hat^2 ).

So the surface area corresponding to our area increment is sqrt( 1 + 4 y_hat^2 ) `dA = sqrt( 1 + 4 y_hat^2 ) `dy `dx.

Our Riemann sum leads us to the integral

int ( int ( sqrt(1 + 4 y^2) dy, 0, 1 - x) dx, 0, 1)

STUDENT QUESTION

OK, except I got (5/6)=0.83 and you got 0.63??!?

INSTRUCTOR RESPONSE

.63 is correct to 2 significant figures.

5/6 can't be right. The integral is way too complicated for a rational-number solution. Square roots, at the very least, would be required.

An antiderivative for the y integral is

ln(sqrt(4y^2 + 1) + 2y)/4 + y* sqrt(4 y^2 + 1)/2.

You get this by starting with the trig substitution y = tan(theta) / 4. You use the identity 1 + tan^2(theta) = sec^2(theta), and y ' = tan(theta) sec(theta) / 4 so dy = 1/4 tan(theta) sec(theta) dTheta. Fairly straightforward, if you're up on trig substitution.

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Self-critique (if necessary):

I got the jist of this, but my answer differed from the given.

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You used f_x and f_y rather than f_x^2 and f_y^2.

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Self-critique rating:OK

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Question: `q002. Find the surface area of the portion of the sphere x^2 + y^2 + z^2 = 36 that lies above the plane z = 4.

The sphere intersects the plane when

x^2 + y^2 + 4^2 = 36 so that

x^2 + y^2 = 20.

This describes a circle of radius 2 sqrt(5) centered at the origin.

This circle can be described by

-2 sqrt(5) <= x <= 2 sqrt(5)

-sqrt(20 - x^2) <= y <= sqrt(20 - x^2)

A vector normal to the sphere at point (x, y, z), obtained using the gradient, is 2 x i + 2 y j + 2 z k. This should be no surprise, since the radial vector x i + y j + z k is normal to the surface at (x, y, z); the first of these is just double the first.

Using either we will find that cos(theta) = z / sqrt(x^2 + y^2 + z^2) = sqrt(36 - x^2 - y^2) / 6, so we will integrate 6 / sqrt( 36 - x^2 - y^2) over our circle..

Alternatively we can solve the equation of the sphere for z, obtaining z = f(x, y) = sqrt( 36 - x^2 - y^2 ). In this case f_x = - x / sqrt( 36 - x^2 - y^2), f_y = - y / sqrt( 36 - x^2 - y^2), and sqrt( 1 + f_x^2 + f_y^2) = 6 / sqrt( 36 - x^2 - y^2).

We integrate the expression sqrt( 1 + f_x^2 + f_y^2) = 6 / sqrt( 36 - x^2 - y^2) over the region, obtaining

int(int(6 / sqrt( 36 - x^2 - y^2) dy, -sqrt(20 - x^2) , sqrt(20 - x^2) ) dx, -2 sqrt(5), 2 sqrt(5) ).

The result is 24 pi.

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Your solution:

x^2 + y^2 + z^2 = 36

z=4

x^2 + y^2 + 4^2 = 36

x^2 + y^2 + 16 = 36

x^2 + y^2 = 20

-sqrt(20 - y^2) <== x <== sqrt(20 - y^2)

-sqrt(20 - x^2) <== y <== sqrt(20 - x^2)

2d circle at origin in xy plane with r=sqrt(20)

I got confused because for your x values you have 2sqrt(5) or sqrt(20). Why isnt it sqrt(20 - y^2)?

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The region in question lies only above the circle x^2 + y^2 = 20.

This region is described by

-2 sqrt(5) <= x <= 2 sqrt(5)

-sqrt(20-x^2) <= y <= sqrt(20 - y^2).

x values move from -2 sqrt(5) to +2 sqrt(5), and for a given x value the y values within the region run from -sqrt(20-x^2) to +sqrt(20-x^2).

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q003. Find the surface area of the portion of the cylinder x^2 + z^2 = 9 which lies above the triangle with vertices (0,0,0), (1,1,0), (1,0,0).

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Your solution:

z^2=(9-x^2)

z=sqrt (9-x^2)

[0<=x<=1], [0<=y<=1-x]

[Int Int ((sqrt 9)/(sqrt (9-x^2) dy, 0, 1-x)dx, 0, 1]

3*[sin^(-1)*(1/3)+2*sqrt(2)-3]

confidence rating #$&*:

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Given Solution:

The cylinder has gradient vector 2 x i + 2 z k, so cos(theta) = 2 z / 2 sqrt( x^2 + z^2) = z / sqrt( x^2 + z^2) and sec(theta) = sqrt( x^2 + z^2 ) / 2 z. Since z = sqrt(9 - x^2) this simplifies to sqrt( 9 ) / ( sqrt( 9 - x^2) ). This is the function we integrate over the region.

Alternatively z = f(x,y) = sqrt( 9 - x^2) so f_x = -x / sqrt( 9 - x^2), f_y = 0 and sqrt( 1 + f_x^2 + f_y^2 ) = sqrt( 1 + x^2 / (9 - x^2) ) = sqrt( 9 / (9 - x^2) ) = 3 / sqrt( 9 - x^2 ).

Either way this function is integrated over the region 0 < x <= 1, 0 <= y <= 1 - x.

The integral is

int( int( 3 / sqrt( 9 - x^2) dy, 0, 1-x) dx, 0, 1)

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Self-critique (if necessary):

OK

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Self-critique rating:OK"

@&

Good work. Check my notes.

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