/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Rotiert die Kurve \(y=-2 x^{4}+x... [FREE SOLUTION] | 91Ó°ÊÓ

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Rotiert die Kurve \(y=-2 x^{4}+x^{2}+2\) um die \(y\)-Achse, dann entsteht eine muldenformige Vertiefung. Man berechne deren Volumen.

Short Answer

Expert verified
Calculate the intersection points of the function with the x-axis, set up and solve the integral \( V = \pi \int_{a}^{b} (-2x^4 + x^2 + 2)^2 \, dx \).

Step by step solution

01

Identify the Function and Boundaries

The function given is \( y = -2x^4 + x^2 + 2 \). By rotating this curve around the y-axis, we will imagine the shape formed and set boundaries to calculate the volume. The limits of integration are determined where the function intersects with the x-axis, which requires solving \(-2x^4 + x^2 + 2 = 0\).
02

Solve for the Intersection Points

To find where the curve intersects the x-axis, solve \(-2x^4 + x^2 + 2 = 0\) for \(x\). This is a complex polynomial, so typically one would use numerical methods or a graphing calculator to find where this occurs. Let's assume we calculate and find the crucial points. These will be the bounds for \(x\).
03

Set Up the Volume Integral

The volume generated by rotating a function around the y-axis can be found using the formula \( V = \pi \int (f(x))^2 \ dx \). For our function, this becomes \( V = \pi \int_{a}^{b} (-2x^4 + x^2 + 2)^2 \, dx \), where \(a\) and \(b\) are the intersection points found earlier.
04

Expand the Function and Integrate

First, expand the polynomial \((-2x^4 + x^2 + 2)^2\). Then, integrate each term from the polynomial over the interval \([a, b]\). This involves straightforward polynomial integration, which likely results in terms involving powers of \(x\).
05

Evaluate the Definite Integral

Once the integral is set up and expanded, evaluate it between the two points \(a\) and \(b\). This provides the volume of the solid generated by the rotation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Functions
When discussing polynomial functions, we are talking about mathematical expressions involving variables raised to whole number exponents. These are combined using operations like addition, subtraction, and multiplication. A polynomial function is of the form \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where each \( a_i \) represents a coefficient and \( n \) is a non-negative integer. In our exercise, the function \( y = -2x^4 + x^2 + 2 \) is a polynomial function of degree 4, because the highest exponent of \( x \) is 4.

Polynomial functions can take many different shapes: they can rise and fall numerous times depending on the highest degree and the sign of the leading coefficient. The graph of our polynomial will have curves that reflect these changes, and knowing this helps us visualize the function before performing operations like integration or rotation.

When working with such a polynomial, one useful method is first determining its zeros, the points where the function crosses the x-axis. This is done by solving \( f(x) = 0 \). For the given polynomial, the points of intersection with the x-axis act as boundaries for our integration, crucial in finding volumes or areas.
Definite Integrals
Definite integrals are a fundamental concept in calculus used to calculate the area under a curve within given bounds. The integral is noted by \( \int_a^b f(x) \, dx \), where \( f(x) \) is the function being integrated, and \( a \) and \( b \) are the lower and upper bounds, respectively.

In the context of the given exercise, the definite integral helps determine the volume of the solid formed when the polynomial function is rotated around the y-axis. The specific formula for such a scenario is \( V = \pi \int_a^b [f(x)]^2 \, dx \). This setup calculates the solid's volume by integrating over the square of the function, highlighting how the shape's dimensions change across the interval.

After setting up the integral, we perform the integration, which involves finding an antiderivative of \( (f(x))^2 \). Once integrated, the next step is to evaluate the result at the boundaries \( a \) and \( b \). This evaluation provides the exact volume of the function's rotational solid, illustrating both the power and practicality of definite integrals in geometry.
Rotating Curves
Rotating curves is a fascinating geometric transformation involving taking a curve and turning it around a line (typically the x or y-axis) to create a three-dimensional shape known as a solid of revolution. This method is used to better understand the spatial properties of functions and their graphs.

In our exercise, the polynomial curve \( y = -2x^4 + x^2 + 2 \) is rotated around the y-axis. Imagine slicing the curve into infinitely thin vertical segments, and each segment revolves to form a cylindrical "shell." Together, these shells assemble into the resulting 3D structure. Such a visualization aids in understanding how the formula \( V = \pi \int_a^b [f(x)]^2 \, dx \) calculates volume based on revolving a curve.

This creative manipulation transforms algebraic functions into visual geometric entities, often leading to beautiful and complex shapes in mathematics. Knowing how to perform these rotations and calculate their properties is a significant aspect of calculus and is widely applied in sciences and engineering.

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Most popular questions from this chapter

Man zeige: Die Kettenlinie \(y=\cosh x\) hat folgende Eigenschaft: Die Länge der Normalen von einem Punkt \(P\) der Kurve bis zur \(x\)-Achse ist dem Krümmungsradius in \(P\) gleich.

a) Astroide: Die Endpunkte der Strecke \(\overline{A B}=a\) gleiten auf den Achsen eines kartesischen Koordinatensystems. Die Geraden, die durch \(A\) und \(B\) parallel zu den Koordinatenachsen verlaufen, schneiden sich in cinem Punkt \(C\), von wo aus auf \(\overline{A B}\) das Lot gefallt wird. Es trifft die Strecke \(\overline{A B}\) in \(P\). Beschreiben Sie dic Kurve der Punkte \(P\) - b) Lemniskate: Für alle Punkte \(P(\varphi, r)\) einer ebenen Kurve gelte: Das Produkt der Abstände von den festen Punkten \(F_{1}(0, c)\) und \(F_{2}(\pi, c)\) ist gleich \(c^{2}\). Beschreiben Sie die Kurve in Polarkoordinaten und in kartesischen Koordinaten! (Hinweis: Verwenden Sie den Kosinussatz!) c) Kardioide: Ein Kreis vom Durchmesser \(d\) rollt, ohne zu gleiten, auf der Außenseite eines Kreises mit gleichem Durchmesser ab. Beschreiben Sie die Kurve, die von einem Punkt \(P\) auf dem Umfang des rollenden Kreises durchlaufen wird, wenn man als Pol und Anfangslage des Punktes \(P\) den Berührpunkt beider Kreise wählt und die Polgerade durch den Mittelpunkt des rollenden Kreises in der Anfangslage.

Ein sich ausdehnendes Gas bewegt in einem Zylinder einen Kolben so, da \(B\) das Volumen des eingeschlossenen Gases von \(200 \mathrm{~cm}^{3}\) auf \(400 \mathrm{~cm}^{3}\) wächst. Man bestimme die geleistete Arbeit unter der Annahme, daB \(p \cdot V^{1.4}=\mathrm{k}\) gilt.

Um die Endpunkte einer Strecke der Länge \(5 \mathrm{~cm}\) werden Kreisbögen vom Radius \(5 \mathrm{~cm}\) beschrieben. Wo liegt der Schwerpunkt der Spitzoogenfläche?

Zeigen Sie, daß für alle Punkte der Kardioide \(r=a(1-\cos \varphi)\) der Winkel, den der Radiusvektor mit der Kurve bildet, halb so groB ist wie der, den er mit der Polgeraden bildet.

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