r/MathHelp • u/Ok-Company282 • 10d ago

Problem with area bounded by polar curves

How do I know when to get a 2 or 4 multiplied before the integral? Youtubers and even my lecturer drew the curves to understand but how do I figure out the shape? Will I even get time to draw the curve in the exam?

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u/The_Card_Player 10d ago

This question doesn't seem very well worded.

If you're evaluating an integral to find area under a real-valued function, any real-valued coefficients in the function you're integrating will likely be written right there in the function definiion. So in that sense you'll know to multiply the 'main' portion of your integral by whatever coefficient happens to be in the function - whether that coefficient is 2, 4, or any other real number.

My best guess is that you might be asking about surface area on some 2D shape created with reflections of a smaler shape in the cartesian plane.

eg 'what is the surface area of a shape enclosed by sin(x) on 0<=x<=pi above and -sin(x) on the same domain below?'

In this case you might compute the area under sin(x) on the given domain and then *double* it to find the total area. This is because the overall shape occupies the integral's area and also the same area mirror flipped over the x axis.

In general the total area of a shape represented by multiple instances of a shape with some intregal-computable area will be the relevant integral value multiplied by however many instances of the smaller shape are included in the larger shape.

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u/cdfe_88 6d ago

the factor depends on the axes of symmetry your shape has. If it only has one, you usually do the integral on one half of the curve and multiply by 2. If it's symmetrical in two axes you integrate in one quadrant and multiply by 4.