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Example Questions
Example Question #2041 :微积分Ii
Let R be the region between f(x) and g(x) on the given interval. FInd the volume V obtained by revolving R around the x-axis.
Interval:
This volume can be found with the washer method, which is done using the equation
where the region to be revolved around the x axis has the upper bound of f(x), the lower bound of g(x), and exists on the interval [a, b]. Applying this equation:
Example Question #2042 :微积分Ii
Let R be the region between the function f(x) and g(x) on the interval given below. Find the volume V obtained by revolving R around the y-axis.
Interval:
This volume can be found using the shell method by implementing the formula
Where the region to be rotated about the y-axis is between f(x) and g(x) on the interval.
Additionally, one can notice that here, because f(x)=-g(x), the region between the two function is the same as twice the region betwee f(x) and the x-axis, simplifying our problem. Using this symmetry and the above formula:
Example Question #2043 :微积分Ii
Using integration by parts, evaluate the following indefinite integral:
In order to evaluate an integral by parts, utilize the equation:
For the integral given, we can define u and dv:
From this, we can evalute du and v:
Combining these elements usng the equation above, we can begin to evaluate the integral:
The final term of this answer needs to be further, again by integration by parts:
Doing the integration:
年代ubstituting this in back for the previous partial solution:
Example Question #121 :Integral Applications
Evaluate the following trigonometric integral:
To solve this, first factor out one cosine and write the rest in terms of sine:
From here, we can do a substitution where u=sin(x) and du=cos(x)dx:
Finally, re-substitute sin(x) for u:
Example Question #11 :Application Of Integrals
Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of
on the interval
units cubed
units cubed
units cubed
units cubed
units cubed
The formula for volume of the region revolved around the x-axis is given as
where
As such
When taking the integral, we use the inverse power rule which states
Applying this rule term by term we get
And by the corollary of the Fundamental Theorem of Calculus
As such the volume is
units cubed
Example Question #10 :Volume Of Cross Sections And Area Of Region
Find the volume of the solid generated by rotating about the y-axis the region under the curve, fromto.
None of the other answers
年代ince we are revolving a function ofaround the y-axis, we will use the method of cylindrical shells to find the volume.
Using the formula for cylindrical shells, we have
.
Example Question #122 :Integral Applications
Find the volume of the solid region swept out by the area bounded by the the functionsandabout the-axis.
Find the volume of the solid region swept out by the area bounded by the the functionsandabout the-axis.
The cross sectional area of the solid can be written as a function ofby first finding the cross-sectional areas swept out by each of the individual functions as they rotate about the x-axis, and then subtracting the inner area.
Visualize a line drawn perpendicularly to the x-axis at some point meeting the function. As we rotate about the x-axis, we sweep out a circular cross-sectional area with a radius equal to the function valueat the corresponding value of. The area of this disk is therefore:
年代imilarly for the function, the cross-sectional area is:
The plot of the two functions helps visualize the geometry. The parabola is clearly the "inner" radius, and therefore we must subtract the areasfromto find the cross-sectional area of the solid:
Now we must find the limits of integration by finding the intersection points.
年代olve for
The other solution is. Now simply integrate the cross sectional areaover the intervalto find the volume of the solid,
To simplify further, write the terms with a common denominator and factor.
Example Question #123 :Integral Applications
Find the volume of solid of revolution for the given function:
.
The volume of a solid of revolution for a given function is found using the disk method. Answer is obtained upon evaluating the following integral:
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