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Homework Help: Hopefully someone will help me - Analysis Problems

  1. Jan 31, 2005 #1


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    1. Find the volume of the solid in [itex]\mathbb{R}^3[/itex] cut out by z = 4x² + y² and z = 4x + 3.

    So I need to find the volume of:

    [tex]A = \{ (x, y, z) \in \mathbb{R}^3 : 4x^2 + y^2 \leq z \leq 4x + 3\}[/tex]

    [tex]v(A) = \int _A 1 = \int _{-0.5} ^{1.5} \int _{-\sqrt{-4x^2 + 4x + 3}} ^{\sqrt{-4x^2 + 4x + 3}} \int _{4x^2 + y^2} ^{4x + 3}\, dz\, dy\, dx[/tex]

    [tex]= \int _{-0.5} ^{1.5} \int _{-\sqrt{-4x^2 + 4x + 3}} ^{\sqrt{-4x^2 + 4x + 3}} (4x + 3 - 4x^2 - y^2)\, dy\, dx[/tex]

    [tex]= \int _{-0.5} ^{1.5} \left (\left ( (4x + 3 - 4x^2)y - \frac{y^3}{3}\right )_{-\sqrt{-4x^2 + 4x + 3}} ^{\sqrt{-4x^2 + 4x + 3}}\right )\, dx[/tex]

    [tex]= \frac{4}{3} \int _{-0.5} ^{1.5} (4x + 3 - 4x^2)^{1.5}\, dx[/tex]

    Is this right so far? If so, how do I solve it?

    2. Give an example of a function [itex]f(x,\, y)[/itex] on [itex][0,\,1] \times [0,\,1] \subset \mathbb{R}^2[/itex] that is (Riemann) integrable on [itex][0,\, 1]\times [0,\,1][/itex] but such that [itex]y \to f(0, y)[/itex] is not (Riemann) integrable on [itex][0,\,1][/itex].

    I chose f(x, y) = 0 if x > 0 or y is rational and 1 otherwise. Then the set {0} x [0, 1] is the set of discontinuities of f, and has 2-dimensional measure 0 (in fact content 0), so f is integrable. However, [itex]y \to f(0, y)[/itex] is continuous on all of [0, 1], which does not have 1-dimensional measure 0, so such a mapping is not integrable.

    3. Let [itex]f[/itex] be (Riemann) integrable on [itex][0,\,1]\times [0,\,1] \subset \mathbb{R}^2[/itex]. Let [itex]B = \{x \in [0,\,1] : y \to f(x,\,y) \mbox{ is not (Riemann) integrable on } [0,\,1]\}[/itex]. Explain why [itex]B[/itex] must be of measure zero.

    If x is in B, then {y in [0, 1] : y --> f(x,y) is discontinuous at y} does not have (1-dimensional) measure 0, and is therefore uncountable. But if y --> f(x, y) is discontinuous at y, then (x, y) --> f(x, y), i.e. f, is discontinuous at y. Therefore, any collection of open rectangles {[itex]U_i[/itex]} covering B has total measure that isn't arbitrarily small. Now, we can choose arbitrarily small open X containing x such that the collection of subsets of [itex]\mathbb{R}^2[/itex] in the form [itex]X \times U_i[/itex] contains some of the points (x, y) for fixed x where f is discontinuous.

    But if B does not have measure zero, it is not countable, thus there are uncountably many x. Therefore, even though the sets covering some of the discontinuities of f for some fixed x have a total volume that is arbitrarily small, it is still positive and finite, and so the total volume of all these sets for all x in B (thus for "most" of the discontinuities of f) is an uncountable sum of arbitrarily small positive finite numbers, so it doesn't converge and this set of discontinuities does not have measure 0. Since this is just a subset of all discontinuities and does not have measure 0, the set of all discontinuities of f does not have measure 0, so f is not integrable. This contradicts the fact that it is, so the bold assumption is false, proving that which was required.

    Is this proof valid?

    4. Let A be a closed rectangle in n-dimensional Euclidean space and f a real-valued bounded function on A. State a theorem which gives a necessary and sufficient condition for f to be integrable on A.

    If and only if the set of discontinuities of f on A has n-dimensional measure 0 is f integrable.

    5. Using the above (or otherwise) prove that if f and g are bounded integrable function on A, then so is their product fg. (Note (fg)(x) = f(x)g(x), not f(g(x)))

    The product of 2 continuous functions is continuous, so fg is discontinuous where f or g are discontinuous. The set of discontinuities of fg will be the union of the set of discontinuities of f and g, but both those sets have measure 0, so their union has measure 0, and fg is hence integrable.

    6. Find the volume of the solid in [itex]\mathbb{R}^3[/itex] which lies above the cone [itex]z^2 = x^2 + y^2[/itex] and inside the sphere [itex]x^2 + y^2 + z^2 = 4z[/itex].

    Rewrite the sphere [itex]x^2 + y^2 + (z-2)^2 = 2^2[/itex] so it is the sphere of radius 2 centered at (0, 0, 2). The volume is:

    [tex]\int _2 ^2 \int _{-\sqrt{4 - x^2}} ^{\sqrt{4 - x^2}} \int _{\sqrt{x^2 + y^2}} ^{2 + \sqrt{4 - x^2 - y^2}}\,dz\,dy\,dx[/tex]

    Is this right so far? If so, is it easy to compute? I tried a while ago, I remember getting stuck.

    7. Find the point on the ellipse [itex]\frac{x^2}{4} + \frac{y^2}{9} = 1[/itex] closest to the point (1, 1).

    The point on the ellipse will be in the 1st quadrant.

    [tex]y = \sqrt{9 - \frac{9}{4}x^2}[/tex]

    [tex]\frac{dy}{dx} = \frac{-2.25x}{\sqrt{9 - \frac{9}{4}x^2}}[/tex]

    It's negative reciprocal is:

    [tex]\frac{\sqrt{9 - \frac{9}{4}x^2}}{2.25x}[/tex]

    This is the slope that the line segment joining (1, 1) and the closest point on the ellipse must have. The slope of the segment joining any point (x, y) on the ellipse will be:

    [tex]\frac{\Delta y}{\Delta x} = \frac{1 - \sqrt{9 - \frac{9}{4}x^2}}{1 - x}[/tex]


    [tex]\frac{1 - \sqrt{9 - \frac{9}{4}x^2}}{1 - x} = \frac{\sqrt{9 - \frac{9}{4}x^2}}{2.25x}[/tex]

    There has to be a better way. I also considered expressing the distance from a point on the ellipse to (1, 1) as a function D of (x, y), but since y is a function of x on the relevant interval, D as a function of x. Then differentiate with respect to x, find the minimum, but that was also ugly. Is there a better way?

    8. Define measure 0 and Jordan measurable and state two theorems that relate these concepts to Riemann integrability.

    A set S has measure 0 if, for any e > 0, we can choose a cover of S by (countably many) open rectangles such that the sum of the volumes of the rectangles is less than e. A set is Jordan measurable if its boundary has measure 0.

    A function f on a closed rectangle is integrable on that rectangle if the set of discontinuities of f on that rectangle has measure 0. If a set is Jordan measurable, then the characteristic function of that set is integrable on a closed rectangle containing that set.
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  3. Feb 1, 2005 #2


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    It looks OK. As for the last integral, 4x+3-4x^2=4(1-(x-0.5)^2). Substitute

  4. Feb 1, 2005 #3


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    For the 6-th problem,check again the limits of integration.With the ones you posted,your result will be 0...

  5. Feb 1, 2005 #4


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    Thanks a lot ehlid, I'm getting [itex]4\pi[/itex]. dextercioby, yeah, typo: it should go from -2 to 2.
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