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Math  /  Calculus

QuestionNow integrate with respect to xx, where x∈[−2,2]x \in[-2,2] : V=∫x=−22[84−x2−2x24−x2−2(4−x2)3/23]dxV=\int_{x=-2}^{2}\left[8 \sqrt{4-x^{2}}-2 x^{2} \sqrt{4-x^{2}}-\frac{2\left(4-x^{2}\right)^{3 / 2}}{3}\right] d x

Studdy Solution

STEP 1

What is this asking? We need to compute a definite integral, which represents a volume VV, over the interval [−2,2][-2, 2]. Watch out! Don't forget to apply the limits of integration correctly after integrating and be careful with the fractional exponents.

STEP 2

1. Simplify the integrand
2. Integrate term by term
3. Evaluate the definite integral

STEP 3

Alright, let's **get this party started**!
First, let's peep that integrand.
It looks a little intimidating, right?
But don't worry, we'll **tame it**!
Notice that each term has a 4−x2\sqrt{4 - x^2} or (4−x2)3/2(4 - x^2)^{3/2}.
Let's **factor that bad boy out** to make things cleaner!

STEP 4

V=∫−224−x2(8−2x2−2(4−x2)3)dx V = \int_{-2}^{2} \sqrt{4 - x^2} \left( 8 - 2x^2 - \frac{2(4 - x^2)}{3} \right) dx See? Much **neater** already!
Now, let's **simplify** the expression inside the parentheses.

STEP 5

V=∫−224−x2(8−2x2−83+2x23)dx V = \int_{-2}^{2} \sqrt{4 - x^2} \left( 8 - 2x^2 - \frac{8}{3} + \frac{2x^2}{3} \right) dx We **distributed** that −23-\frac{2}{3}.
Now **combine like terms**!

STEP 6

V=∫−224−x2(24−83−6x2−2x23)dx V = \int_{-2}^{2} \sqrt{4 - x^2} \left( \frac{24 - 8}{3} - \frac{6x^2 - 2x^2}{3} \right) dx V=∫−224−x2(163−4x23)dx V = \int_{-2}^{2} \sqrt{4 - x^2} \left( \frac{16}{3} - \frac{4x^2}{3} \right) dx Almost there!
Let's **factor out** 43\frac{4}{3}.

STEP 7

V=43∫−224−x2(4−x2)dx V = \frac{4}{3} \int_{-2}^{2} \sqrt{4 - x^2} (4 - x^2) dx V=43∫−22(4−x2)3/2dx V = \frac{4}{3} \int_{-2}^{2} (4 - x^2)^{3/2} dx Boom! **Way easier** to handle, right?

STEP 8

Now, this integral screams **trig substitution**!
Let x=2sin⁡(θ)x = 2\sin(\theta), so dx=2cos⁡(θ)dθdx = 2\cos(\theta) d\theta.
When x=−2x = -2, sin⁡(θ)=−1\sin(\theta) = -1, so θ=−π2\theta = -\frac{\pi}{2}.
When x=2x = 2, sin⁡(θ)=1\sin(\theta) = 1, so θ=π2\theta = \frac{\pi}{2}.

STEP 9

Also, 4−x2=4−4sin⁡2(θ)=4(1−sin⁡2(θ))=4cos⁡2(θ)4 - x^2 = 4 - 4\sin^2(\theta) = 4(1 - \sin^2(\theta)) = 4\cos^2(\theta).
So, (4−x2)3/2=(4cos⁡2(θ))3/2=8cos⁡3(θ)(4 - x^2)^{3/2} = (4\cos^2(\theta))^{3/2} = 8\cos^3(\theta).

STEP 10

Substituting, our integral becomes: V=43∫−π/2π/28cos⁡3(θ)⋅2cos⁡(θ)dθ=643∫−π/2π/2cos⁡4(θ)dθ V = \frac{4}{3} \int_{-\pi/2}^{\pi/2} 8\cos^3(\theta) \cdot 2\cos(\theta) d\theta = \frac{64}{3} \int_{-\pi/2}^{\pi/2} \cos^4(\theta) d\theta Much better!

STEP 11

Now, we can use the **reduction formula** for cos⁡n(θ)\cos^n(\theta) or the **double angle formula** for cos⁡2(θ)\cos^2(\theta) repeatedly.
Let's use the latter: cos⁡2(θ)=1+cos⁡(2θ)2 \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} So, cos⁡4(θ)=(1+cos⁡(2θ)2)2=14(1+2cos⁡(2θ)+cos⁡2(2θ)) \cos^4(\theta) = \left( \frac{1 + \cos(2\theta)}{2} \right)^2 = \frac{1}{4} (1 + 2\cos(2\theta) + \cos^2(2\theta)) cos⁡4(θ)=14(1+2cos⁡(2θ)+1+cos⁡(4θ)2)=18(3+4cos⁡(2θ)+cos⁡(4θ)) \cos^4(\theta) = \frac{1}{4} \left( 1 + 2\cos(2\theta) + \frac{1 + \cos(4\theta)}{2} \right) = \frac{1}{8} (3 + 4\cos(2\theta) + \cos(4\theta))

STEP 12

V=643∫−π/2π/218(3+4cos⁡(2θ)+cos⁡(4θ))dθ=83∫−π/2π/2(3+4cos⁡(2θ)+cos⁡(4θ))dθ V = \frac{64}{3} \int_{-\pi/2}^{\pi/2} \frac{1}{8} (3 + 4\cos(2\theta) + \cos(4\theta)) d\theta = \frac{8}{3} \int_{-\pi/2}^{\pi/2} (3 + 4\cos(2\theta) + \cos(4\theta)) d\theta V=83[3θ+2sin⁡(2θ)+14sin⁡(4θ)]−π/2π/2 V = \frac{8}{3} \left[ 3\theta + 2\sin(2\theta) + \frac{1}{4}\sin(4\theta) \right]_{-\pi/2}^{\pi/2}

STEP 13

**Time to plug and chug** those limits! V=83[(3π2+0+0)−(−3π2+0+0)]=83⋅3π=8π V = \frac{8}{3} \left[ \left( \frac{3\pi}{2} + 0 + 0 \right) - \left( -\frac{3\pi}{2} + 0 + 0 \right) \right] = \frac{8}{3} \cdot 3\pi = \mathbf{8\pi}

STEP 14

The volume VV is 8π8\pi.

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