How do you find the area of a region using integrals with respect to x or y?

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Homework Statement


decide whether to integrate with respect to x or y. draw a typical approximating rectangle and label its height and width then find the aread of the region

X^2 = y, x-2y = 3

The Attempt at a Solution


i'm not sure how to find if the questions should be repsect to x or y
i tried doing it with respect to X I did it with integral with upper bound being 9 and lower 0 and i plugged in rad x + (x-3)/2 dx and got 45 -4 and the answer is 32/3
HELP
 
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Did you draw a graph of the region? The graphs of the functions you gave don't intersect, so they don't determine a region, so you're going to have a tough time finding its area.

Have you posted the right equations?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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