# Thread: A few Calc II Final Exam Questions (Setup ONLY)

1. ## A few Calc II Final Exam Questions (Setup ONLY)

Just a few setup problems... If I knew what strategy to use to evaluate these integrals, I'd be better off. That's all I need to know for the following:

1.) Find the volume of the solid formed by revolving the region bounded by the graphs of $\displaystyle y = -x^2 + 4$ and y = 0 about the x-axis.

2.) Evaluate:

$\displaystyle \int_{-4}^4 \sqrt{16 - x^2}dx$

3.) Evaluate:

$\displaystyle \int \ln{2x} \ dx$

2. Originally Posted by Aryth
Just a few setup problems... If I knew what strategy to use to evaluate these integrals, I'd be better off. That's all I need to know for the following:

1.) Find the volume of the solid formed by revolving the region bounded by the graphs of $\displaystyle y = -x^2 + 4$ and y = 0 about the x-axis.

2.) Evaluate:

$\displaystyle \int_{-4}^4 \sqrt{16 - x^2}dx$

3.) Evaluate:

$\displaystyle \int \ln{2x} \ dx$
For the first problem, use the disc method and integrate:
$\displaystyle \int_{-2}^{2} \pi{(4 - x^2)^2} dx$

3. Thanks, that helped tremendously.

4. 2. $\displaystyle y = + \sqrt{16 - x^{2}}$
$\displaystyle y^{2} = 16 - x^{2}$
$\displaystyle x^{2} + y^{2} = 16$

The integral represents the area under the upper half of a circle of radius 4. So: $\displaystyle \text{A} = \frac{1}{2}\pi r^{2}$

3. $\displaystyle u = 2x \: \Rightarrow \: du = 2 dx \: \Rightarrow \: dx = \frac{du}{2}$

Making the subs:
$\displaystyle \int \ln u \: \frac{du}{2} = \frac{1}{2} \int \ln u \: du$
which is a standard one

5. I'm not sure about #2, but for #3 the antiderivative of $\displaystyle \ln{x}$ is $\displaystyle x \ln{x} - x + C$.

6. Originally Posted by o_O
2. $\displaystyle y = + \sqrt{16 - x^{2}}$
$\displaystyle y^{2} = 16 - x^{2}$
$\displaystyle x^{2} + y^{2} = 16$

The integral represents the area under the upper half of a circle of radius 4. So: $\displaystyle \text{A} = \frac{1}{2}\pi r^{2}$
Wow, thanks. That was completely different then what I was expecting. But the answer was correct.

7. Originally Posted by Aryth
Wow, thanks. That was completely different then what I was expecting. But the answer was correct.
Yeah, I tried using trig substitution on that one. Not the way to go about it, apparently.

8. Well you could do a trig sub:
$\displaystyle x = 4\sin \theta$
$\displaystyle dx = 4\cos \theta d\theta$

So you get:
$\displaystyle = \int \sqrt{16 - 16\sin^{2} \theta} \cdot 4\cos \theta d\theta$
$\displaystyle = 4 \int \sqrt{16} \sqrt{1 - \sin^{2} \theta} \cos \theta d\theta$

etc. etc.

But as you can see having that little insight with recognizing the form of a circle speeds up the process.