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Find the area between $y=9\sin x\,$ and $\,y = 10\cos x\,$ over the interval $[0, \pi]$.

Having a tough time with this problem. Im good with finding the area between two curves but the sin and cos threw me off a lot.

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  • $\begingroup$ Have you tried graphing or drawing it to get an idea of what it looks like? $\endgroup$ – ferson2020 Feb 6 '13 at 20:22
  • $\begingroup$ okay thank you. sorry $\endgroup$ – Ak47 Feb 7 '13 at 0:01
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You'll want to find the point of intersection, and use the $t=x$-value of this point as a bound for integrating (see note below).

Set the two equations equal to one another, and solve for the values $t$ (the x-coordinate) that solves the resulting equation: there will be one such value on your domain.

$$9\sin(t) = 10\cos(t)\implies \frac{\sin t}{\cos t} = \frac {10}{9}\implies \tan t = \frac{10}{9} \implies t = \tan^{-1} \frac{10}{9}$$

Then calculate the sum of the integrals $$\int_0^t (10\cos x - 9\sin x)\,dx + \int_t^\pi (9\sin x- 10\cos x)\,dx.$$


Note: you need to know the point of intersection to divide the integral into two parts, because from $0\leq x \lt t$, where t is the value of x at the point of intersection, $10\cos x > 9\sin x$. At $x = t$, $10\cos x = 9 \sin x$. And from $t \lt x \leq \pi$, $9\sin x > 10 \cos x$. So to measure total area, we need to ensure we choose the correct function as the "upper bound" of each region when we integrate to obtain the total area of the regions bound by the functions.

enter image description here

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  • $\begingroup$ Great. You shouldn't call $x$ $x$, though, since it is also your integration variable. $\endgroup$ – Julien Feb 6 '13 at 20:59
  • $\begingroup$ thanks @julien, good point. $\endgroup$ – Namaste Feb 6 '13 at 21:02
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Hint: the curves look like this:

enter image description here

You need to find the point of intersection and split the interval into two pieces.

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As mrf suggested. Find the $x$ value at the intersection of the two curves. Then calculate the integral $\int_0^x(upper curve) - (lower curve)\,\mathrm{d}x$ + $\int_x^\pi(upper curve) - (lower curve)\,\mathrm{d}x$.

$$\int_0^x(10 \cos(x) - 9\sin(x))\,\mathrm{d}x + \int_x^\pi(9\sin(x) - 10 \cos(x))\,\mathrm{d}x$$

You should end up with $2\sqrt{181}$which is approximately 26.907.

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