# What are the correct ways to use polar form in solving integrals?

• transgalactic
In summary, for the first problem, the given equations x= a cos(\alpha), y= a sin(\alpha) do not represent the astroid curve in polar form. The correct polar form is x= r cos(\alpha), y= r sin(\alpha), where both r and \alpha are independent variables. The parametric equations for the astroid curve are x= a sin3(t), y= a cos3(t) and the arc length is given by ds= \sqrt{9a^2 cos^2(t)sin^3(t)+ 9a^2 sin^2(t) cos^4(t)}dt. For the second problem, the correct polar coordinates are x= r cos\theta
For the first problem, find the length of the "astroid", x2/3+ y2/3[/sup]= a2/3, you say you wanted to put it in "polar form" and then give x= a cos(($\alpha$), y= a sin($\alpha$). That is NOT "polar form. Polar form would be x= r cos($\alpha$), y= r sin($\alpha$), where both r and $\alpha$ are independent variables. There is no reason to thing that x= a cos(($\alpha$), y= a sin($\alpha$) satisfy the equation of the curve.

As for the "method" your text uses, you say they let x= a sin3(t), y= a cos3(t). Okay, then x2/3+ y2/3= a2/3sin2(t)+ a2/3cos2(t)= a2/3 so those are parametric equations for the curve. Also, then dx= 3a cos(t)sin(2(t)dt and dy= -3a sin(t)cos2(t) dt so that ds= $\sqrt{(dx/dt)^2+ (dy/dt)^2}= \sqrt{9a^2 cos^2(t)sin^3(t)+ 9a^2 sin^2(t) cos^4(t)}dt$.

As for the second one, y2= x2(a2- x2, i don't see why you again have "x= a cos$\alpha$, y= a sin$\alpha$". Polar coordinates are x= r cos$\theta$, y= r sin[\theta]. (of course, it doesn't matter whether you use $\alpha$ or $\theta$. $\theta$ is the standard notation.)

Then $y^2= r^2 sin^2(\theta)$ and $x^2= r^2 cos^2(\theta)$ so your equation becomes $r^2 sin^2(\theta)= r^2(a^2- r^2 cos^2(\theta)$. We can cancel the two $r^2$ terms and then we have $sin^2(\theta)= a^2- r^2 cos^2(\theta)$ so $r^2 cos^2(\theta)= a^2- sin^2(\theta)$, $r^2= a^2 sec^2(\theta)- tan^2(\theta)$.

## 1. What is the purpose of using integrals in scientific research?

Integrals are used to find the area under a curve, which can be applied in many scientific fields such as physics, engineering, and economics. They also allow us to calculate important quantities such as volume, displacement, and probability.

## 2. How do you solve a definite integral?

To solve a definite integral, you first need to evaluate the indefinite integral of the function. Then, you plug in the upper and lower limits of integration and subtract the result at the upper limit from the result at the lower limit. This will give you the exact value of the definite integral.

## 3. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration and gives a single numerical value as the result. An indefinite integral does not have any limits and gives a general expression in terms of a constant. It represents the antiderivative of a function.

## 4. How do you interpret the area under a curve on a graph?

The area under a curve on a graph represents the total accumulation of the function over the given interval. This can have various interpretations depending on the context, such as total displacement or total profit.

## 5. What are some real-world applications of integrals?

Integrals have many real-world applications, such as calculating the area of irregular shapes, determining the distance traveled by an object with varying velocity, and finding the volume of objects with irregular shapes. They are also used in fields such as engineering, economics, and physics to solve various problems and make predictions.

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