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polar system

b0t2

New member
Oct 26, 2013
1
I have a task to solve it in polar system: x²=4y-y²; x²=8y-y²; y=x; x=0. So in polar: r=4sin phi; r=8sin phi; phi=pi/4; phi=pi/2. The integral - int(from pi/4 to pi/2) d phi int(from 4 sin phi to 8 sin phi) r dr. My answer is 3pi - 1/4 but seems like its not true. Somebody has another answer?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I have a task to solve it in polar system: x²=4y-y²; x²=8y-y²; y=x; x=0. So in polar: r=4sin phi; r=8sin phi; phi=pi/4; phi=pi/2. The integral - int(from pi/4 to pi/2) d phi int(from 4 sin phi to 8 sin phi) r dr. My answer is 3pi - 1/4 but seems like its not true. Somebody has another answer?
Welcome to MHB, b0t2! :)

Let's rewrite your statements in $\LaTeX$.
Apparently your problem has:
$$x^2=4y-y^2; \quad x^2=8y-y^2; \quad y=x; \quad x=0$$
So in polar:
$$r=4\sin \phi; \quad r=8\sin \phi; \quad \phi=\pi/4; \quad \phi=\pi/2$$
The integral:
$$\int_{\pi/4}^{\pi/2} d \phi \int_{4 \sin \phi}^{8 \sin \phi} r dr$$
Your answer is $3\pi - 1/4$ which appears not to be true.

Perhaps you can clarify?
From your first statement I deduce x=y=0, but obviously that is not what you intended.
I suspect you're talking about some kind of intersection of surfaces, but I prefer not to guess.
Do you perhaps have the complete problem statement?