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 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 xaxis.
 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 xaxis.

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shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.1x = 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. \[\int\limits_{0}^{5} \pi*(5^2  x^2)dx + \int\limits_{0}^{\sqrt(5)}\pi*(5^2  x)dx\]

Cameronmx9
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhh there are two shapes..

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.01st 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 \[V = \pi \times[\int\limits_{5}^{0} (x  5)^2 dx + \int\limits_{0}^{\sqrt{5}} (x^2 5)^2 dx]\]

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.1@campbell ... it won't be (x5)^2 but (x^2  5^2) ...

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.1think ... equation of volume between two cylinders is not (r1  r2)^2 but (r1^2  r2^2)

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.1int\limits_{0}^{5} \pi*(5^2  x^2)dx + \int\limits_{0}^{\sqrt(5)}\pi*(5^2  x^4)dx would be the answer...

Cameronmx9
 3 years ago
Best ResponseYou've already chosen the best response.0What 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 yaxis, rotated along the following lines: x=36

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.1find 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 \[\int\limits_{0}^{6} \pi*(36^2  (y^4))dy\]

Cameronmx9
 3 years ago
Best ResponseYou've already chosen the best response.0So this integral would = 1296y  y^5/5?
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