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For , √3 tan $=1/153 T3 @= tan F6 xl P: IT 2 ^¹ (73) FAISzer 2²=3y² 2=√By Z= 7:mx ====√ √ 2 Y tan (3 G

For , √3 tan $=1/153 T3 @= tan F6 xl P: IT 2 ^¹ (73) FAISzer 2²=3y² 2=√By Z= 7:mx ====√ √ 2 Y tan (3 G

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.3: Volume And Average Value
Problem 23E
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Please explain the part circles in pink and also explain how you know that you should do this part circled in pink.

√3 4. Integral Calculus of Multivariable Functions 4. Let E be the ice cream cone solid, convert to spherercal bounded above by x² + y² + (2-1)² = 1 and bounded below by the cone z2 = 3(x² + y²). Find the volume of E. 2²=3(x² + y²) (2-1)(2-1) x² + y ² +(²-1) ² = 1 For 4, x² + y² + z ²2²-22+1=1 p²-2pros4=0 P= 2 cos & √3 Z tan $=1/1535 = tan an" (T3) P= I b ✓ x² + y² + (2-1)² = 1 p=21us & y 2= √√3y tan = FXIS zero 2²=3y² 2= √3y 7:mx LIT I 6 2005 √ √ So p²sing apdødo O Jan (G) tar 1-2 > Y √3 (2-1)(2-1) 2²-2211 x² te 4.7: Spherical Coordinates ry² + 2²--22+1 2103& Jo Jo S p²sin & dp d øde 0 p²- - 2 pous = 0 P(P-2 cos): 0 p=0 and p2105p
Transcribed Image Text:√3 4. Integral Calculus of Multivariable Functions 4. Let E be the ice cream cone solid, convert to spherercal bounded above by x² + y² + (2-1)² = 1 and bounded below by the cone z2 = 3(x² + y²). Find the volume of E. 2²=3(x² + y²) (2-1)(2-1) x² + y ² +(²-1) ² = 1 For 4, x² + y² + z ²2²-22+1=1 p²-2pros4=0 P= 2 cos & √3 Z tan $=1/1535 = tan an" (T3) P= I b ✓ x² + y² + (2-1)² = 1 p=21us & y 2= √√3y tan = FXIS zero 2²=3y² 2= √3y 7:mx LIT I 6 2005 √ √ So p²sing apdødo O Jan (G) tar 1-2 > Y √3 (2-1)(2-1) 2²-2211 x² te 4.7: Spherical Coordinates ry² + 2²--22+1 2103& Jo Jo S p²sin & dp d øde 0 p²- - 2 pous = 0 P(P-2 cos): 0 p=0 and p2105p
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