Volume of a solid y=cos⁡(x) and y=0 for the interval 0≤x≤π2 I used method of shells to get: 2π∫0π2xcos⁡(x)dx And I got: 2π∫0π2xcos⁡(x)dx=2π[xsin⁡(x)+cos⁡(x)]0π2 =2π[π2sin⁡(π2)−cos⁡(0)] =2π(π2−1) 2π∫0π2xcos⁡(x)dx=π2−2π Then I am doing it by method of slicing and got this as an answer

You are correct. Using the distribute law (⋆), we have 2π(π2−1)=(⋅)(2π)(π2)−(2π)(1)=π2−2π

Expert Answer to "Volume of a solid y=cos⁡(x) and y=0 for..."

Hailee Cline

2022-01-24

Volume of a solid

y = \cos (x)

and y=0 for the interval

0 \leq x \leq \frac{π}{2}

I used method of shells to get:

2 π \int_{0}^{\frac{π}{2}} x \cos (x) d x

And I got:

2 π \int_{0}^{\frac{π}{2}} x \cos (x) d x = 2 π {[x \sin (x) + \cos (x)]}_{0}^{\frac{π}{2}}

= 2 π [\frac{π}{2} \sin (\frac{π}{2}) - \cos (0)]

= 2 π (\frac{π}{2} - 1)

2 π \int_{0}^{\frac{π}{2}} x \cos (x) d x = π^{2} - 2 π

Then I am doing it by method of slicing and got this as an answer

sineurosi0f

Beginner2022-01-25Added 14 answers

You are correct.
Using the distribute law (⋆), we have

2 π (\frac{π}{2} - 1) \overset{(\cdot)}{=} (2 π) (\frac{π}{2}) - (2 π) (1) = π^{2} - 2 π

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