please solve all 34. -1 points LarCalc10 11.8.035. points LarCalc10 11 Find an e

Question

please solve all

34. -1 points LarCalc10 11.8.035. points LarCalc10 11 Find an equation for the surface of revolution formed by revolving the curve in the indicated coordinate plane about the given axis. Equation of Curve xy = 9 Coordinate Plane xy-plane Axis of Revolution 35. -1 points LarCalc10 11.6.038. Find an equation for the surface of revolution formed by revolving the curve in the indicated coordinate plane about the given axis Equation of Cure Coordinate Plane Axis of Revolution x=ln z a-plane 36. -1 points LarCalc10 121.002. Find the domain of the vector-valued function. (Enter your answer using interval notation.)

Explanation / Answer

Let P(x, y, z) be a generic point on the surface of revolution. Fix a point Q(x, y1, 0) on the curve with the same y-coordinate as P. Then we have

x*y1 = 9

The square of the distance from Q to the x-axis is y1^2
and the square of the distance from P to the
x-axis is y^2 + z^2

As the two distances should be the same, we have
y1^2 = y^2 + z^2

Therefore, the
wanted equation can be obtained by replacing y1^2 :

x*y1 = 9

So, x^2*y1^2 = 81

x^2 * (y^2 + z^2) = 81

So, we have
x^2y^2 + x^2z^2 = 81 -----> ANS

--------------------------------------------------------------------

35)
x = ln(z)

Let P(x, y, z) be a generic point on the surface of revolution. Fix a point Q(x, 0, z1) on the curve with the same x-coordinate as P. Then we have

x = ln(z1)

The square of the distance from Q to the x-axis is z1^2
and the square of the distance from P to the
x-axis is y^2 + z^2

As the two distances should be the same, we have z1^2 = y^2 + z^2

Now, replacing into x = ln(z) :
x = ln(sqrt(y^2 + z^2))

x = 1/2 * ln(y^2 + z^2) ----> ANS

--------------------------------------------------------------------

36)
r = <sqrt(25-t^2) , t^2 , -8t>

t^2 and -8t can have any value of t subb-ed into them...

But sqrt(25 - t^2)
We need
25 - t^2 >= 0

t^2 - 25 <= 0

(t + 5)(t - 5) <= 0

This satisfies
[-5 , 5]

So, domain is :
t belongs to [-5 , 5] ----> ANS

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.