# Matematik fvvdfsdc

y = \frac {3x} {\sqrt {x^2+1}} \text{ eğrisi } x=0 \text{ ve } x=8 \text{ doğruları ve } x \text{ ekseni ile sınırlanan alanın } x \text{ ekseni etrafında döndürülmesiyle oluşturulan döner cisimin hacmini bulunuz.

Seçenekler:

1. ( \frac{3\pi}{2} \ln(50) )
2. ( \frac{3\pi}{2} \ln(37) )
3. ( \frac{3\pi}{2} \ln(65) )
4. ( \frac{3\pi}{2} \ln(50) )
5. ( \frac{3\pi}{2} \ln(65) )

Solution:

To find the volume of the solid formed by rotating the given curve ( y = \frac {3x} {\sqrt {x^2+1}} ) around the ( x )-axis from ( x = 0 ) to ( x = 8 ), we will use the disk method, where the volume is given by:

V = \pi \int_{a}^{b} [y(x)]^2 \, dx
1. Determine ( [y(x)]^2 ):
y = \frac{3x}{\sqrt{x^2+1}} \implies y^2 = \left( \frac{3x}{\sqrt{x^2+1}} \right)^2 = \frac{9x^2}{x^2+1}
1. Set up the integral:
V = \pi \int_{0}^{8} \frac{9x^2}{x^2+1} \, dx
1. Simplify the integral:

Let us use the substitution method to solve the integral. Setting ( u = x^2 + 1 ) gives ( du = 2x , dx ) or ( \frac{1}{2} du = x , dx ).

When ( x = 0 ), ( u = 1 ) and
when ( x = 8 ), ( u = 64 + 1 = 65 ).

Hence, the integral becomes:

V = 9\pi \int_{1}^{65} \frac{x^2}{x^2+1} \, dx = 9\pi \int_{1}^{65} \frac{u-1}{u} \cdot \frac{1}{2} du = \frac{9\pi}{2} \int_{1}^{65} \left(1 - \frac{1}{u}\right) du
1. Integrate:
V = \frac{9\pi}{2} \left[ u - \ln|u| \right]_{1}^{65}

Evaluating this gives:

V = \frac{9\pi}{2} \left[ (65 - \ln 65) - (1 - \ln 1) \right]

where ( \ln(1) = 0 ):

V = \frac{9\pi}{2} (65 - \ln 65 - 1)
V = \frac{9\pi}{2} (64 - \ln 65)

Therefore, the correct answer matches option III or V:

V = \frac{3\pi}{2} \ln(65) \text{ (this matches both III and V)}