# Integral - I can't get the right answer

• Sane
In summary, the conversation discusses finding the volume of a solid obtained when a given region is rotated about the x-axis. The formula for the area of the sample is simplified to A(x) = pi(sin(x))^3 and integrated between 0 and pi. The correct way to integrate is by splitting (sin(x))^3 as (sin(x))^2*sin(x). The integral of (sin(x))^2 can be solved using substitution, while the integral of sin(x) involves the quotient rule. Therefore, the correct formula for the volume is V = pi * (4/3) = (4/3)pi.
Sane
Find the volume of the solid obtained when the given region is rotated about the x-axis.

Under $y = (\sin{x})^{\frac{3}{2}}$ between 0 and pi.

The radius is... $r = (\sin{x})^{\frac{3}{2}}$

Then the area for any sample is... $A (x) = \pi((\sin{x})^{\frac{3}{2}})^{2}$

Simplifying to... $A (x) = \pi(\sin{x})^{3}$

Integrate between 0 and pi to get the volume...

$$V = \pi \int_{0}^{\pi} (\sin{x})^{3}$$

$$V = \pi [ \frac{(\sin{x})^{4}}{4\cos{x}} ]_{0}^{\pi}$$

But... sin(pi) and sin(0) both equal 0, making the volume 0. But it's actually (4/3)(pi). What am I missing?

Sane said:
Integrate between 0 and pi to get the volume...

$$V = \pi \int_{0}^{\pi} (\sin{x})^{3}$$

$$V = \pi [ \frac{(\sin{x})^{4}}{4\cos{x}} ]_{0}^{\pi}$$

That's not the right way to integrate...

hint: Split (sin(x))^3 as (sin(x))^2.sin(x)

You are missing the fact that
$$\int(\sin{x})^{3}dx$$
is NOT
$$\frac{(\sin{x})^{4}}{4\cos{x}}$$
Since there was not a cosine in the original integrand, you can't simply put one in. In particular,if you try to differentiate
$$\frac{(\sin{x})^{4}}{4\cos{x}}$$
you do not get $sin^3(x)$ because you have to use the quotient rule and differentiate that new "cos(x)" also.

You surely learned how to do integrals of odd powers of sine and cosine long ago: separate out one "sin x" to use with the derivative:
[tex]\int sin^2(x) sin(x)dx= \int (1- cos^2(x)) sin(x)dx[/itex]
and let u= cos(x).

(Blast you, neutrino, you beat me again!)

Last edited by a moderator:
I think you got confused doing the integral.

It's true that Integral sin^3x cos x dx = 1/4 sin^4 x

But you can't just "divide by cos x" and say Integral sin^3 x = 1/4 sin^4 x / cos x. Check out the ways you have learned to do integrals like sin^3 x.

Ahhh shoot! It didn't even strike me to use substitution. I was being ignorant to the fact that it would require the quotient rule to derivate what I thought was the integral... Thank-you very much! I'm so relieved!

Sane said:
it would require the quotient rule to derivate what I thought was the integral

The word is 'differentiate'. :)

## What is Integral?

Integral is a mathematical concept that represents the area under a curve on a graph. It is typically used to solve problems involving rates of change or accumulation.

## Why can't I get the right answer with Integral?

There are several reasons why you may not be getting the correct answer when using Integral. This could be due to inputting incorrect values, not using the correct formula, or making a mistake during the calculation process.

## How do I solve a problem using Integral?

To solve a problem using Integral, you first need to identify the function or curve for which you need to find the area under. Then, use the appropriate formula to calculate the integral with the given values. Finally, double check your work to ensure accuracy.

## What are some common mistakes to avoid when using Integral?

Some common mistakes to avoid when using Integral include inputting incorrect values, using the wrong formula, forgetting to change the limits of integration, and making calculation errors. It is important to check your work carefully to catch and correct any mistakes.

## Can I use Integral to solve any type of problem?

Integral can be used to solve problems involving rates of change or accumulation, but it may not be the most efficient or appropriate method for every problem. It is important to understand when and how to use Integral in order to effectively solve problems.

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