(a) find the volume of the solid obtained by revolving the region bounded by the

Question

(a) find the volume of the solid obtained by revolving the region bounded by the curves y= 2(x)^.5, y=0 and x=4 about the line y= 2

(b) find the surface area of the surface generated by revolving the curve y=2(x)^0.5, 0<=x<=4, about the line y=0.

Explanation / Answer

Points of intersection: x^(1/2) = x^2 ==> x = x^4 (with x non-negative, since x^(1/2) is undefined for negative reals) ==> x(x^3 - 1) = 0 ==> x = 0 or 1; the other roots are not real. Hence the points of intersection are (0, 0) and (1, 1). Drawing a picture will be useful for below. ---------------------------------- I'll do Cylindrical Shells first. For x in (0, 1), we have x^(1/2) > x^2 (try x = 1/4 for example). Hence, V = (x = 0 to 1) 2(x - 0)(x^(1/2) - x^2) dx, A = 2rh for a cylindrical shell ...= (x = 0 to 1) 2 (x^(3/2) - x^3) dx ...= 2 ((2/5) x^(5/2) - (1/4)x^4) {for x = 0 to 1} ...= 3/10. ------------------------ For Disk/Washer (slicing), we need to rewrite the equations in terms of y (since we're revolving around the y-axis). y = x^(1/2) ==> x = y^2 y = x^2 ==> x = y^(1/2). Since x = y^(1/2) is to the right of x = y^2 for y in (0, 1), V = (y = 0 to 1) [(y^(1/2))^2 - (y^2)^2] dy, via A = r^2 for a disk ...= (y = 0 to 1) (y - y^4) dy ...= (y^2/2 - y^5/5) {for y = 0 to 1} ...= 3/10. I hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.