Pls, help me with this problem
Calculate the volume of the body formed by rotating a figures (x^2)+3y=0 and y+3=0 around the Y-axis.
Thanks!
First thing to do is calculate your limits of integration. You can do that by finding where your curves meet.
In the second equation, you can see that y = -3 so substitute y = -3 into the first equation and solve for x.
x^2 + 3*-3 = 0
x = (+/-)3, so we're integrating from x=-3 to 3.
Since our limits of integration are in terms of x, we will have to use a method that uses an integral with respect to x.
Rotating about the y axis & integrating with respect x means we'll be using the shell method.
Now we know what our integral will look like: 2(pi)*INTEGRAL(r(x)h(x)dx) (from a to b)
where r(x) is the distance from the axis of rotation and h(x) is the height at x.
Because our axis is the y-axis, distance from it is simply 'x.'
For h(x) we have to ask our self which curve is on top, and we observe that x^2 - 9 < -3 on the interval [-3, 3]. so our height function will be
h(x) = (-3) - (x^2 - 9).
Now we know all the parts needed to calculate volume:
V = 2(pi)*INTEGRAL(x*[(-3) - (x^2 - 9)]dx) from -3 to 3.
Look this picture
I think that it is better to shift the body of rotation for three units up along the Y-axis, i.e. in the new coordinate system $\displaystyle {x^*} = x
$ and $\displaystyle {y^*} = y + 3$ we have $\displaystyle y_1^* = {y_1} + 3 = - 3 + 3 = 0$ and $\displaystyle y_2^* = {y_2} + 3 = - \frac{{{x^{*2}}}}{3} + 3 = 3 - \frac{{{x^{*2}}}}{3}$.
Then $\displaystyle {V_y} = 2\pi \int\limits_0^3 {{x^*}\left( {3 - \frac{{{x^{*2}}}}{3}} \right)d{x^*}} = \ldots = \frac{{27}}{2}\pi {\text{ }}\left( {{\text{cubic units}}} \right).$