Here are two questions:
- Find the volume of the solid of revolution, generated by rotating the region bounded by the graph of $y = \arcsin x$ and the lines $x = 1$ and $y = 0$ through 2$\pi$ radians about the y axis.
- Find the volume of a solid that is obtained by rotating the curve $y = \arccos x$, $0 ≤ x ≤ 1$, through $2 \pi$ radians about the y-axis.
Now, I am new to integration but I decided to give this a go.
My method was as follows:
Question 1 - integrate $\pi(\sin y)^2$with respect to y and evaluate at the limits 0 and 1.
Using this method, I got an answer of $\frac12$ - $\frac14 4\sin(2)$. This is wrong, however.
I used the same method for Question 2 but again, got an answer different to the solutions given.
Would anyone be able to help me in order to determine the correct method for this problem and also such problems in general?
Here is a working for the first one. I will leave for you to do the second one using similar approach.
The shaded region in the diagram is bound by the curve $y = \arcsin (x), y = 0$ and $x=1$. This shaded region is being revolved around y-axis and we need to find the volume of the solid of revolution. If we are applying washer method,
The lower and upper bound of radial distance from y-axis is $x = \sin y$ and $x = 1$. Bound of $y$ is between $0$ and $\frac{\pi}{2}$.
So $V = \displaystyle \int_0^{\pi/2} \int_{\sin y}^1 2 \pi x \ dx \ dy = \pi \int_0^{\pi/2} (1-\sin^2y) \ dy = \frac{\pi^2}{4}$