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AP Calculus BC : Volume of Cross Sections and Area of Region

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Example Question #181 : Ap Calculus Bc

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

g(t)=4sin(t)-6cos(t)

Possible Answers:

2.59

11.07

-1.41

-2.59

Correct answer:

2.59

Explanation:

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

g(t)=4sin(t)-6cos(t)

We are asked to find the area under a curve. This sounds like a job for an integral.

We need to integrate our function from 0 to $t=\frac{3 \pi}{4}$ . These will be our limits of integration. With that in mind, our integral look like this.

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt$

Now, we need to recall the rules for integrating sine and cosine.

$1) \int sin(t)dt=-cos(t)+c$

$2) \int cos(t)dt=sin(t)+c$

With these rules in mind, we can integrate our original function to get:

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

$G(t)=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

Now, we just need to evaluate our new function at the given limits. Let's start with 0

G(0)=-4cos(0)-6sin(0)+c=-4-0+c=-4+c

And our upper limit...

$G({\frac{3 \pi}{4}})=-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c$

$-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c=(-4)\frac{-\sqrt{2}}{2}-(6)\frac{-\sqrt{2}}{2}+c$

$(-4)\frac{-\sqrt{2}}{2}-(6)\frac{\sqrt{2}}{2}+c=2\sqrt{2}-3\sqrt{2}+c=-\sqrt{2}+c$

Now, we need to take the difference between our limits:

$G(\frac{3 \pi}{4})-G(0)=(-\sqrt{2}+c)-(-4+c)=-\sqrt{2}+4\approx2.585$

So, our answer is 2.59

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AP Calculus BC : Volume of Cross Sections and Area of Region

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Example Question #181 : Ap Calculus Bc

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