## Cameronmx9 3 years ago *CALC2 HELP* The region enclosed by the curve x=sqrt(y), and by the lines x= -y and y=5 rotated around the x-axis.

1. shaan_iitk

x = sqrt(y) and x = -y only intersect on (0,0) x = sqrt(y) and y = 5 intersect on (sqrt(5), 5) x = -y and y = 5 intersect on (-5,5) Now we have two volumes when we rotate about x axis --> 1st volume for the triangle formed in the second coordinate and the other for the parabola in the 1st quadrant. $\int\limits_{0}^{5} \pi*(5^2 - x^2)dx + \int\limits_{0}^{\sqrt(5)}\pi*(5^2 - x)dx$

2. Cameronmx9

Ohhh there are two shapes..

3. campbell_st

1st rewrite the equation $x = \sqrt{y}$ as y = x^2 where x >=0 to the right of x = 0 the bounded area is x^2 - 5 the point of intersection of y = x^2 and y = 5 is $(\sqrt{5} ,5)$ to the left of x = 0 the bounded area is x - 5 then the volume becomes $V = \pi \times[\int\limits_{-5}^{0} (x - 5)^2 dx + \int\limits_{0}^{\sqrt{5}} (x^2 -5)^2 dx]$

4. shaan_iitk

@campbell ... it won't be (x-5)^2 but (x^2 - 5^2) ...

5. shaan_iitk

think ... equation of volume between two cylinders is not (r1 - r2)^2 but (r1^2 - r2^2)

6. shaan_iitk

int\limits_{0}^{5} \pi*(5^2 - x^2)dx + \int\limits_{0}^{\sqrt(5)}\pi*(5^2 - x^4)dx would be the answer...

7. Cameronmx9

What about this one.. **Calc2 Help Needed** Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=6 , below by the curve y=sqrt(x), and to the left by the y-axis, rotated along the following lines: x=36

8. shaan_iitk

find the intersection of y = 6 and y = sqrt(x) which is (36,6) .. Hence you have to find (area of rectangle with dimensions 36,6 - area bounded by the parabola) Answer would be $\int\limits_{0}^{6} \pi*(36^2 - (y^4))dy$

9. Cameronmx9

So this integral would = 1296y - y^5/5?