Calculus Help: Find Answers to Questions on Volume & Integration

• Bazzdc89
In summary, the first problem is that the student substituted u = x-1 which resulted in the wrong antiderivative. The second problem is that the student is doing the washer method twice which is causing the problem.
Bazzdc89

Hello everyone, first time poster here. Well here is my problem , I recently had a test and right after the test my teacher handed our class the solutions to the test ,there were two problems that my answers conflicted with his. So basically my questions are why are my answers different from his? Am I doing my problems correctly?. I would greatly appreciate it if you could review my answers

$$\int \frac{1}{(x-1)\sqrt{x^2-2x}} \, dx$$

The Attempt at a Solution

so I started with

$$u=x^2-2x\text$$
$${du}=2(x-1)\text{dx}$$
$$\frac{\text{du}}{2}=(x-1)\text{dx}$$

then i let$$u=x-1$$
so $$u+1=x$$

that got me
$$\frac{1}{2}\int \frac{1}{(u+1)\sqrt{u}} \, du$$

I then changed that too
$$\frac{1}{2}\int \frac{1}{u^{\frac{3}{2}}+\sqrt{u}} \, du$$

then I let
$$v=\sqrt{u}$$
$$v^2=u$$

$$2\text{vdv}=\text{du}$$

$$\frac{2}{2}\int \frac{v}{\left(v^2\right)^{\frac{3}{2}}+\sqrt{v^2}}$$
Simplifying I got
$$\int \frac{1}{v^2+1} \, dv$$
Then from there it was easy
$$\text{ArcTan}[v]$$
$${ArcTan}\left[\sqrt{u}\right]$$
$${ArcTan}\left[\sqrt{x^2-2x}\right]$$

so that was my final answer
Here is my second question with deals with volume of a solid under revolutionFind the volume of the solid generated by revolving the region bounded by the graphs of
y=sin(x) y=cos(x) x= Pi/2 ( we could use which ever method we preferred, I used washer) My attempt Using Washer method
a=0 , b= pi/4
$$\int \text{Cos}[x]^2-\text{Sin}[x]^2dx$$
Minus

a= pi/4 , b= pi/2
$$\int \text{Sin}[x]^2-\text{Cos}[x]^2dx$$which resulted my in getting
$$\pi$$

My teachers answers to the two questions
For the first one
$$\sec ^{-1}(x-1)$$
For the second one
$$\left(\frac{\pi }{2}-1\right)\pi$$

Last edited:
Your attempt at the first problem is wrong because you have u = x^2 - 2x and u = x-1. You can't define u to be two different expressions and then substitute u back in.

So I completely forgot how to calculate volumes, but in the washer method don't you subtract two integrals that have the same limits of integration?

I noticed that I did use two U substitutions, but maple and wolfram are giving me the same answer so I'm kind of confused on that first step of substitution.

And for washer you do the integration of from a to b, "Big radius"- "small radius" , squaring each of the radius,and for my problem i had to do the washer method twice cause the graph of cos(x)>sin(x),from 0-->Pi/4 ; and the graph of sin(x) > cos(x) from Pi/4 --> Pi/2

Well the issue is that you're basically treating those two expressions to be the same. Not only that, but technically you made a mistake in substituting u since you put u = x-1, so you should end up with u^(3/2) in the denominator (temporarily allowing the substitution). Thus I think it was by coincidence that whatever calculations you made resulted in the correct antiderivative. The point is you can't just let u be two different things. The solution given by wolfram alpha is completely different, and does not let u be two different expressions.

Alright so I figured out what my problem was for question #1, It was coincidence how I got the right answer, and I figured out the proper way to solve the integration.

Any idea if I am solving question #2 right?

1. What is the purpose of volume and integration in calculus?

The purpose of volume and integration in calculus is to calculate the total amount of space or quantity within a given region or function. Integration is used to find the area under a curve, which can then be used to calculate volume by rotating the curve around an axis. These concepts are important in many areas of science and engineering, such as physics, economics, and engineering.

2. How do you find the volume of a 3-dimensional object using calculus?

To find the volume of a 3-dimensional object using calculus, you can use the formula V = ∫a^b A(x) dx, where A(x) represents the cross-sectional area of the object at a given value of x. This means that you need to find the function that represents the cross-sectional area and then integrate it over the desired interval. This process can be applied to a wide range of 3-dimensional objects, such as cylinders, cones, and spheres.

3. Can calculus be used to find the volume of irregularly shaped objects?

Yes, calculus can be used to find the volume of irregularly shaped objects. This is done by breaking down the object into smaller, simpler shapes, such as rectangles or triangles, and then using integration to find the volume of each shape. The volumes of these smaller shapes can then be added together to find the total volume of the irregularly shaped object.

4. What is the relationship between volume and integration?

The relationship between volume and integration is that integration is used to find the volume of 3-dimensional objects. By integrating a function that represents the cross-sectional area of an object, you can find the total volume of that object. This relationship is based on the fundamental theorem of calculus, which states that integration and differentiation are inverse operations.

5. How can calculus be used to solve real-life problems involving volume and integration?

Calculus can be used to solve real-life problems involving volume and integration by providing a mathematical tool to analyze and solve complex problems. For example, in engineering, calculus can be used to calculate the volume of a water tank or the amount of material needed to construct a bridge. In economics, calculus can be used to optimize profits by finding the volume of production that maximizes revenue. In physics, calculus can be used to calculate the volume of an irregularly shaped object, such as a planet or asteroid. These are just a few examples of how calculus can be applied to real-life problems involving volume and integration.

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