Paremetric surface revolved around y-axis

Question

if I'm finding the area of the surface generated by revolving the curve around the y-axis I use the equation $2\pi x\sqrt{(x')^2+(y')^2}$ and I'm given $$x=(2/3)t^{3/2}$$ $$y=2\sqrt{2}$$ and I got this was derivatives $$x'=t^{1/2};y'=t^{-1/2}$$ that gives me $$(2\pi(2/3)t^{3/2})\sqrt{ ((t^{-1/2})^2)+t }$$ then I u sub gives.. $$\pi(2/3)u^{3/2}$$I go from $$0<t<\sqrt{3}$$ I get a negative number. what am I doing wrong? du is 2x =$$2(2/3)t^{3/2}$$ right $$u= (t^{-1/2})^2)+t $$

Answer 1

This is a simplified version of the integral you need to find. All I have done is write things like $(t^{-1/2})^2 = t^{-1}$ as $\frac{1}{t}$. I am omitting the constant out in front as well. Multiply by it at the end.

$\int_0^{\sqrt{3}} t^{3/2} \sqrt{\frac{1}{t}+t} dt.$ Consider letting $u = t^2$. Then $du = 2t$. This is not the most intuitive $u$ substitution, but it makes things much cleaner.

Looking at just the pieces of the integrand: $t^{3/2} \sqrt{\frac{1}{t}+t} = t\sqrt{t}\sqrt{\frac{1}{t}+t} = t\sqrt{t(\frac{1}{t}+t)} = t\sqrt{1+t^2}$.

We then make the u substitution. Since $du = 2t dt$, we divide by 2. The $t$ outside the square root is taken care of by replacing $t~dt$ with $du$.

We are left with the integral $\dfrac{1}{2}\displaystyle\int \sqrt{u + 1}~ du = \dfrac{1}{3}(u + 1)^{3/2}.$

So, back substituting yields $\dfrac{1}{3} (t^2+1)^{3/2}.$

Evaluating at 0 to $\sqrt 3$ gives $\dfrac{1}{3} (3+1)^{3/2} - \dfrac{1}{3} (1)^{3/2} = \dfrac{7}{3}.$

Multiplying by our previous constant 2π(2/3), we get $\dfrac{28\pi}{9}$.